Current Electricity - Class 12 Physics - Chapter 3 - Notes, NCERT Solutions & Extra Questions
Renews every month. Cancel anytime
Your personal doubt-solving assistant
Chatterbot AI gives you 100% accurate answers to your questions in an instant.
Extra Questions - Current Electricity | NCERT | Physics | Class 12
What effect does adding lightbulbs have in a series circuit? Describe how the voltage changed across the different points in a series circuit.
When additional lightbulbs are added to a series circuit, several key changes occur, especially in terms of voltage distribution and power dissipation across each bulb, which in turn affects brightness.
Voltage Distribution Across Bulbs:
In a series circuit, the total voltage supplied by the source is divided equally among all the bulbs assuming they have equal resistance. Consider a scenario with three bulbs connected in series to a voltage source $ V $:
If each bulb has resistance $ R $, the total resistance $ R_{\text{eq}} $ of the circuit is the sum of the resistances of each bulb. So, $ R_{\text{eq}} = R + R + R = 3R $.
The current $ I $ flowing through the circuit can be expressed as: $$ I = \frac{V}{R_{\text{eq}}} = \frac{V}{3R} $$
Voltage across each bulb then becomes: $$ V_{\text{AB}} = V_{\text{BC}} = V_{\text{CD}} = I \times R = \frac{V}{3} $$
If an additional bulb is added, making it four bulbs in series: $$ R_{\text{eq}} = R + R + R + R = 4R $$
The new current $ I' $ would be: $$ I' = \frac{V}{4R} $$
Voltage across each bulb now: $$ V'{\text{AB}} = V'{\text{BC}} = V'{\text{CD}} = V'{\text{DE}} = I' \times R = \frac{V}{4} $$
Power Dissipation and Brightness:
The power dissipated in each bulb, which relates to its brightness, is given by $ P = I^2 \times R $.
Initially with three bulbs: $$ P = \left(\frac{V}{3R}\right)^2 \times R = \frac{V^2}{9R} $$
After adding the fourth bulb: $$ P' = \left(\frac{V}{4R}\right)^2 \times R = \frac{V^2}{16R} $$
Significant Observation: The power dissipation per bulb $ P' $ is less than $ P $. Hence, the brightness of each bulb decreases as more bulbs are added to the series.
In summary, adding bulbs to a series circuit results in a reduction of both the voltage drop across each bulb and the brightness of each bulb due to lower power dissipation. The overall resistance of the circuit increases, which in turn reduces the current flowing through each bulb, consequently decreasing the voltage across and the power dissipated by each bulb.
Heating in a current-carrying conductor is a product of _____ and time.
A. current
B. resistance
C. power
Correct Answer: C. power
The heating effect $H$ in a current-carrying conductor is determined by the power $P$ dissipated through the conductor and the time $t$ for which it is applied.
Given the resistance $R$ and voltage $V$, the relationship can be expressed as: $$ H = I^2 \times R \times t = \frac{V^2}{R} \times t $$ where $P = I^2 \times R$ or $P = \frac{V^2}{R}$. Thus, $$ \Rightarrow H = P \times t $$ This demonstrates that the heating in a conductor over a period of time is directly proportional to the power consumed and the duration of power application.
In a series LCR circuit, $L = 10 \ \mathrm{mH}$, $C = 1 \ \mu \mathrm{F}$, and $R = 3 \ \Omega$. It is connected to an AC voltage source. The inductive reactance of the circuit at a frequency $25 %$ less than the resonant frequency is:
A) $75 \ \Omega$
B) $100 \ \Omega$
C) $125 \ \Omega$
D) $150 \ \Omega$
The correct answer is A) $75 \ \Omega$.
To solve this, first calculate the resonant frequency $ \omega_r $, which is given by: $$ \omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10 \times 10^{-3} \times 1 \times 10^{-6}}} = \frac{1}{\sqrt{10^{-8}}} = 10^4 \ \text{rad/s} $$
Since the frequency is given as $25%$ less than the resonant frequency, we calculate $75%$ of $ \omega_r $: $$ \omega = 0.75 \cdot \omega_r = 0.75 \cdot 10^4 = 7500 \ \text{rad/s} $$
We determine the inductive reactance $ X_L $ using: $$ X_L = \omega L = 7500 \cdot 10 \times 10^{-3} = 75 \ \Omega $$
Thus, the inductive reactance of the circuit at the given frequency is $75 \ \Omega$.
Consider the circuit shown. If all the cells have negligible internal resistance, what will be the current through the 2Ω resistor when steady state is reached?
A) 0.66 A
B) 0.29 A
C) OA
D) 0.14 A
The correct answer is C) 0 A.
In the given scenario, the electromotive forces (emfs) of two of the cells negate each other due to their arrangements. When the circuit reaches steady state, particularly considering a capacitor in the circuit, the behavior of the circuit can be simplified for analysis.
At steady state, the capacitor acts as an open circuit for direct current (DC), which means it does not allow any direct current to pass through. Therefore, no current can flow through the 2Ω resistor when the circuit has reached steady state.
Conclusion: The current through the 2Ω resistor is zero amperes.
"Difference between resistance and resistivity."
Resistivity is characterized as the inherent material property representing its ability to impede electric current flow. This property is represented by an SI unit of $\Omega$-m (ohm-meter), highlighting that resistivity is inherent to the material itself and does not depend on the object's dimensions.
The formula for resistance $R$ in terms of resistivity is given as:
$$ R = \rho \frac{\text{length} (L)}{\text{area} (A)} $$
where $\rho$ represents resistivity, $L$ is the length of the conductor, and $A$ is its cross-sectional area. Resistance, on the other hand, greatly depends on both the material properties and the dimensions of the conductor.
Descriptively, resistance is the property that obstructs the flow of electric current, analogous to how a narrow pipe restricts water flow. The voltage drop across a conductor is comparable to the pressure needed to push water through a pipe. The more narrow or longer the pipe (or wire), the greater the resistance to flow. In a cylindrical conductor, for example, resistance $R$ increases with length $L$, and decreases with an increase in cross-sectional area $A$, expressed as:
$$ R \propto \frac{L}{A} $$
Resistivity, $\rho$, thus defines how easily charges flow through a material without considering dimensions, making it an intrinsic property. Resistance $R$, in contrast, varies with the size and shape of the object, classifying it as an extrinsic property. This distinction mirrors the relationship between heat capacity and specific heat in thermodynamics, where one is dependent on mass (extrinsic) and the other is independent of it (intrinsic).
In essence, while both resistance and resistivity relate to how a material opposes electrical current, resistivity is a fundamental property of the material, whereas resistance depends on both the material and its geometric dimensions.
Two metal wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the conductivities of the metals respectively, the effective conductivity of the combination is
A) $\sigma_{1}+\sigma_{2}$
B) $\frac{\sigma_{1}+\sigma_{2}}{2}$
C) $\sqrt{\sigma_{1} \sigma_{2}}$
D) $\frac{\sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$
The correct option for the effective conductivity of two metal wires connected in series, with conductivities $\sigma_1$ and $\sigma_2$, is D) $\frac{\sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}$.
Derivation:
When two resistors are in series, their total resistance $R$ is the sum of their individual resistances $R_1$ and $R_2$: $$ R = R_1 + R_2 $$
The resistance of a material can be expressed in terms of its resistivity $\rho$ and cross-sectional area $A$: $$ R = \frac{\rho L}{A} $$ where $L$ is the length of the wire.
For identically dimensioned wires connected in series, the total resistivity $\rho$ of the combination would also be the sum: $$ \rho = \rho_1 + \rho_2 $$
Since conductivity $\sigma$ is the reciprocal of resistivity ($\sigma = \frac{1}{\rho}$), the combined conductivity $\sigma$ can be expressed as: $$ \frac{1}{\sigma} = \frac{1}{\sigma_1} + \frac{1}{\sigma_2} $$
Solving for $\sigma$, we have: $$ \sigma = \frac{\sigma_1 \sigma_2}{\sigma_1 + \sigma_2} $$
Thus, the effective conductivity of the series combination of two wires with conductivities $\sigma_1$ and $\sigma_2$ is D) $\frac{\sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$. This form is very similar to the parallel resistance formula, highlighting the reciprocal relationship between resistivity and conductivity.
Determine the current in each branch of the network shown in Fig. 3.20:
It seems we encountered issues with directly solving the set of equations. I will break down the solving process step-by-step manually.
Given the equations:
$10 I_1 + 5 (I_1 - I_2) + 10 I_3 = 10$
$5 I_2 + 5 (I_2 - I_1) + 10 I_2 = 0$
$I_1 = I_2 + I_3$
Let's simplify them:
Simplifying the Equations
$10 I_1 + 5 I_1 - 5 I_2 + 10 I_3 = 10$ [ 15 I_1 - 5 I_2 + 10 I_3 = 10 ]
$5 I_2 + 5 I_2 - 5 I_1 + 10 I_2 = 0$ [ 20 I_2 - 5 I_1 = 0 ]
$I_1 = I_2 + I_3$
Solving the Equations
From the second equation: [ 20 I_2 - 5 I_1 = 0 \implies I_1 = 4 I_2 \quad \text{(Eq. 4)} ]
Substitute Eq. 4 into Eq. 1: [ 15 (4 I_2) - 5 I_2 + 10 I_3 = 10 ] [ 60 I_2 - 5 I_2 + 10 I_3 = 10 ] [ 55 I_2 + 10 I_3 = 10 \implies 11 I_2 + 2 I_3 = 2 \quad \text{(Eq. 5)} ]
Using Eq. 3 and Eq. 4: [ 4 I_2 = I_2 + I_3 \implies 3 I_2 = I_3 \quad \text{(Eq. 6)} ]
Substituting Eq. 6 into Eq. 5: [ 11 I_2 + 2 (3 I_2) = 2 ] [ 11 I_2 + 6 I_2 = 2 ] [ 17 I_2 = 2 ] [ I_2 = \frac{2}{17} ]
From Eq. 6: [ I_3 = 3 I_2 = 3 \times \frac{2}{17} = \frac{6}{17} ]
From Eq. 4: [ I_1 = 4 I_2 = 4 \times \frac{2}{17} = \frac{8}{17} ]
Solution
$I_1 = \frac{8}{17} \text{ A}$
$I_2 = \frac{2}{17} \text{ A}$
$I_3 = \frac{6}{17} \text{ A}$
Thus, the currents in each branch of the network are:
Current through branch AB $(I_1)$: $ \frac{8}{17} \text{ A}$
Current through branch BC $(I_2)$: $ \frac{2}{17} \text{ A}$
Current through branch AC $(I_3)$: $ \frac{6}{17} \text{ A} $
"The SI unit of electromotive force is:
A. volt
B. newton
C. joule
D. watt"
The correct answer is A. volt.
The electromotive force (emf) refers to the potential difference created by a source of electrical energy (such as a battery) and is essential for maintaining a continuous flow of current through a conductor. This potential difference becomes evident particularly when no current is being drawn from the battery.
The standard unit of measurement for electromotive force is the volt. Thus, option A is correct.
3 Resistors, one of them having a resistance of $12 \Omega$, when connected in parallel together across a battery of $44 \mathrm{V}$ draws a current of $22 \mathrm{~A}$. These resistors, when connected in series across the same battery, draw a current of $2 \mathrm{~A}$. Find the resistance value of the other 2 resistors.
A. 8,3 ohm
B. 6,4 ohm
C. 12,2 ohm
D. 5,5 ohm
We need to find the values for two unknown resistances when a $12 \Omega$ resistor and these two resistors are connected:
In parallel, drawing a current of $22 , \text{A}$ from a $44 , \text{V}$ battery.
In series, drawing a current of $2 , \text{A}$ from the same battery.
Step 1: Analyzing the Parallel Connection
The total resistance $R_p$ when the resistors are in parallel can be calculated using the formula for parallel resistors: $$ \frac{1}{R_p} = \frac{1}{12} + \frac{1}{x} + \frac{1}{y} $$ Using Ohm’s law $R_p = \frac{V}{I}$, we have: $$ R_p = \frac{44 , \text{V}}{22 , \text{A}} = 2 \Omega $$ So, $$ \frac{1}{12} + \frac{1}{x} + \frac{1}{y} = \frac{1}{2} $$ Then we'll find the net resistance in terms of $x$ and $y$: $$ \frac{1}{2} = \frac{12x + 12y + xy}{12xy} $$ Rearranging terms: $$ 12xy = 2(12x + 12y + xy) $$ This simplifies to: $$ 10xy = 24(x + y) $$
Step 2: Analyzing the Series Connection
When connected in series, the total resistance $R_s$ is: $$ R_s = 12 + x + y $$ Again using Ohm's law $V = IR$, we get: $$ 44 , \text{V} = 2 , \text{A} \cdot R_s \implies R_s = 22 \Omega $$ So, $$ 12 + x + y = 22 \Rightarrow x + y = 10 $$
Matching Choices with Equations
Both equations: $$ 10xy = 24(x + y) $$ and: $$ x + y = 10 $$ need to be satisfied by the given options. The only option matching this criteria is Option B which suggests the resistances are $6 \Omega$ and $4 \Omega$. Verifying this:
Sum is $4 + 6 = 10$.
For the product equation $10 \cdot 6 \cdot 4$ should be equal to $24 \cdot (6 + 4)$: $$ 10 \cdot 24 = 24 \cdot 10 $$ Both equal to $240$, thus confirming that this option fulfills both conditions.
Thus, the correct answer Option B with resistances $6 \Omega$ and $4 \Omega$ is confirmed.
In a hydrogen-oxygen fuel cell, electricity is produced. In this process $\mathrm{H}_{2}(\mathrm{g})$ is oxidised at the anode and $\mathrm{O}_{2}(\mathrm{g})$ is reduced at the cathode. Given: Cathode: $\mathrm{O}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) + 4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-}(\mathrm{aq})$ Anode: $\mathrm{H}_{2}(\mathrm{g}) + 2 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) + 2 \mathrm{e}^{-}$ $4.48 \mathrm{L}$ of $\mathrm{H}_{2}$ at $1 \mathrm{atm}$ and $273 \mathrm{K}$ is oxidised in $9650$ sec. The current produced is (in amps):
A. 1
B. 4
C. 2
D. 8
The correct answer is Option B: 4.
Key principles involved:
Faraday's Law of Electrolysis, which relates the amount of substance reacted or produced to the electric charge passed through: $$ m = z \times i \times t $$ where:
$m$ is the mass of the reactant reacted or product formed.
$z$ is the electrochemical equivalent of the substance.
$i$ is the current in amperes.
$t$ is the time in seconds.
Conversion of gas volume to moles, assuming the gas behaves ideally: The number of moles of a gas can be calculated using the volume of the gas at STP (Standard Temperature and Pressure) divided by the molar volume at STP (22.4 L/mol): $$ \text{number of moles} = \frac{V_{\text{at STP}}}{22.4} $$ Thus: $$ m = \frac{V_{\text{at STP}}}{22.4} \times \text{Molar Mass} $$
Application of these principles:
Given, $V_{\text{H}_2(\text{g})} = 4.48 , \text{L}$ at STP, we calculate the mass of $\mathrm{H}_2$ using the molar mass of $\mathrm{H}_2 (2 , \text{g/mol})$: $$ m = \frac{4.48 , \text{L}}{22.4 , \text{L/mol}} \times 2 , \text{g/mol} $$
Further simplifying and applying Faraday's law: $$ \frac{4.48}{22.4} \times 2 = z \times i \times 9650 $$
Here, we know that the quantity $\frac{4.48}{22.4} \times 2$ evaluates to the total charge corresponding to the reaction, and $t = 9650$ sec. By substituting and simplifying, we find: $$ i = 4 , \text{A} $$ Thus, the system produces a current of 4 amps, making Option B the correct answer.
What is the working principle of an electrical fuse?
A) Ohm's law
B) Joule's law of heating
C) Coulomb's law
D) Faraday's law
The correct answer is B) Joule's law of heating.
An electrical fuse operates based on the heating effect of electric current, which is explained by Joule’s law of heating. A fuse is typically integrated into an electrical circuit in series with the device it protects. The material of the fuse is chosen specifically from metals or alloys with suitable melting points, such as aluminum, copper, iron, or lead. When the current exceeds a pre-determined safe level, the increased current causes the temperature of the fuse wire to rise. This increase in temperature leads to the melting of the fuse wire, consequently breaking the circuit and preventing further damage or hazard.
Which among the following is/are conducting in nature?
A) Human body
B) A bird
C) A copper rod
D) Rubber tyre
The materials that act as conductors possess free electrons which facilitate the flow of electric current. The correct choices that reflect conducting materials in the given list are:
A) Human body
B) A bird
C) A copper rod
The human body and birds, like many living organisms, contain fluids and minerals which can conduct electricity, thus making them conductive. Copper, a metal known for its excellent electrical conductivity, understandably is also a conductor.
On the other hand, materials like rubber tires lack free electrons, making them non-conductive, and are termed as bad conductors or insulators. They do not allow the smooth passage of electrons through them, hindering any electric current flow.
In Ohm's experiment, the reading of the voltmeter is $5.0 \pm 0.2$ and the reading of the ammeter is $3.00 \pm 0.06$. Find the percentage error in $R$.
A) $4 %$
B) $3 %$
C) $6 %$
D) $8 %$
To determine the percentage error in the resistance $R$ measured through Ohm's Law ($R = \frac{V}{I}$), using a voltmeter and ammeter, let's analyze the given errors in their respective readings.
Variance in Voltmeter Reading: The voltmeter has a value $V = 5.0 \pm 0.2$. The percentage error for the voltmeter can be calculated as follows: $$ \text{Error in voltmeter reading} = \frac{\Delta V}{V} \times 100 = \frac{0.2}{5.0} \times 100 = 4% $$
Variance in Ammeter Reading: The ammeter indicates $I = 3.00 \pm 0.06$. Its percentage error is: $$ \text{Error in ammeter reading} = \frac{\Delta I}{I} \times 100 = \frac{0.06}{3.00} \times 100 = 2% $$
Combination of Errors for Derived Measurements: Knowing the rules for propagated errors, if quantities are involved in division or multiplication, the relative errors add up. Thus, the percentage error in resistance $R$ is the sum of the percentage errors of $V$ and $I$: $$ \frac{\Delta R}{R} \times 100 = \frac{\Delta V}{V} \times 100 + \frac{\Delta I}{I} \times 100 = 4% + 2% = 6% $$
Therefore, the correct answer is: C) (6% )
Ohm's law is the relation between the power used, the current, and potential difference.
A) True
B) False
The correct answer is B) False.
Ohm's law specifically establishes a relationship between the current ($I$) flowing through a conductor and the potential difference ($V$) across it. The equation for Ohm's law is represented as: $$ V = IR $$ where $V$ is the voltage in volts, $I$ is the current in amperes, and $R$ is the resistance in ohms. It does not directly relate to power, which involves not only current and voltage but also the factor of energy conversion rate, expressed differently.
Current is due to:
A) charge flowing through the conductor.
B) temperature of the conductor.
C) resistance of the conductor.
D) weight of the conductor.
The correct answer is A) charge flowing through the conductor.
Current is described as the flow of electric charge across a specified area of a conductor in a given period of time. The more technically precise definition is:
$$ \text{Current} (I) = \frac{\text{Charge} (Q)}{\text{Time} (t)} $$
This simply means that current measures how much electric charge passes through the conductor per unit of time. Thus, options B, C, and D are not directly related to the fundamental cause of current.
A thin uniform wire $AB$ of length $50$ cm and resistance $1 \Omega$ is connected to the terminals of a battery of emf $\varepsilon_{1}=2.2$ V and internal resistance $0.1 \Omega$. If the terminals of another cell (assume ideal) are connected to two points $25$ cm apart on the wire $AB$ without altering the current in the wire $AB$, then the emf $\varepsilon_{2}$ of the cell in volts is:
A $0.5$ V
B $1$ V
C $1.2$ V
D $0.8$ V
To determine the emf $\varepsilon_2$ of the ideal cell connected to the wire $AB$, we first calculate the current in the wire $AB$ when it is connected to the battery with emf $\varepsilon_1$ and internal resistance.
The total resistance of the circuit is: $$ R_{\text{total}} = R_{\text{internal}} + R_{AB} = 0.1 , \Omega + 1 , \Omega = 1.1 , \Omega $$
Applying Ohm's Law, the total current ($I$) flowing through the circuit is given by: $$ I = \frac{\varepsilon_1}{R_{\text{total}}} = \frac{2.2 \text{ V}}{1.1 \Omega} = 2 \text{ A} $$
This same current flows through wire $AB$.
Next, we calculate the resistance of the $25$ cm segment of the wire $AB$. Since the wire is uniform and $50$ cm in length corresponds to $1 , \Omega$ resistance, a $25$ cm length will have a resistance of: $$ R_{25 \text{ cm}} = \frac{25 \text{ cm}}{50 \text{ cm}} \times 1 \Omega = 0.5 \Omega $$
Given that no additional current is drawn by the ideal cell, the voltage across this $25$ cm segment is computed as: $$ V = IR_{25 \text{ cm}} = 2 \text{ A} \times 0.5 \Omega = 1 \text{ V} $$
Thus, the emf $\varepsilon_2$ of the ideal cell, required to maintain the existing current without altering it, matches the voltage across the $25$ cm segment: $$ \boldsymbol{\varepsilon_2 = 1 \text{ V}} $$
The correct answer is Option B: $1$ V.
Electrons move at a speed of about $1 \text{ mm} / \text{s}$. Then how does the light turn on as soon as we switch it on?
Think of this scenario as similar to a water pipe connected to a tank. When the tap is opened, water immediately starts coming out at the opposite end because of the pressure in the system, even though the individual water molecules don't travel instantly from the tank to the faucet.
Similarly, electric current behaves like the flow of water. The electrons in a wire move very slowly (at about $1 \text{ mm/s}$), but they create a chain of interactions along the wire almost instantaneously. When a switch is flipped on, these interactions propagate through the circuit at a significant fraction of the speed of light, allowing the electric current to be established immediately. Thus, the bulb glows the moment we turn the switch on, not because the electrons have traveled all the way from the source but because the electrical effect propagates rapidly through the circuit.
Which of the following currents are warm currents?
A Gulf Stream
B Labrador Current
C Oyashio Current
D Kuroshio Current
The Gulf Stream and the Kuroshio Current are both warm currents. Specifically:
A) Gulf Stream: This is a well-known warm ocean current that originates in the Gulf of Mexico and flows into the Atlantic Ocean.
D) Kuroshio Current: This is another warm current, which flows northward, past the eastern coast of Japan.
Conversely, the Labrador Current and the Oyashio Current are cold currents, thus not listed among the warm currents.
If the resistance of a conductor is kept constant and the voltage through the appliance is reduced to half. Then, the resultant power is
(A) $\frac{1}{2} \times P$
(B) $\frac{1}{4} \times P$
(C) $4 \times P$
(D) $\frac{1}{8} \times P$
The correct option is (B) $\frac{1}{4} \times P$.
Given:
The initial power is $P$.
The resistance of the conductor is $R$.
The initial voltage is $V$.
The power formula using resistance and voltage is: $$ P = \frac{V^2}{R} $$
If the voltage is reduced to half, the new voltage, $V'$, is: $$ V' = \frac{V}{2} $$
Substituting $V'$ into the power formula, we find the new power, $P'$: $$ P' = \frac{(\frac{V}{2})^2}{R} = \frac{V^2}{4R} $$
Since $P = \frac{V^2}{R}$, we can rewrite $P'$ as: $$ P' = \frac{1}{4} \times P $$
Thus, when the voltage is halved while keeping the resistance constant, the resultant power is a quarter ($\frac{1}{4}$) of the original power.
According to the convention, the direction of flow of electric current is taken from the positive terminal to the negative terminal of the battery, which is opposite to the flow of...
According to electrical convention, the direction of the electric current is taken opposite to the flow of electrons. Thus, the direction of conventional current flows from the positive terminal to the negative terminal of a battery.
How does a battery create a potential difference in a circuit?
A battery in a circuit creates a potential difference, which essentially is the variation in the energy levels of electrons between two points. Here is how it works:
The negative terminal of the battery is more densely packed with electrons that possess higher energy.
Conversely, the positive terminal contains electrons with lower energy despite the common misconception that it doesn’t have electrons at all. It's crucial to understand that electrons are omnipresent; both terminals of a battery have electrons but with different energy levels.
When a circuit is completed, a potential difference is established between these two terminals — the side with higher energy electrons (negative) and the side with lower energy electrons (positive). This difference drives electrons to flow from the negative side, where energy is higher, towards the positive side where energy is lower, establishing current flow in the circuit.
The internal resistance of a cell of EMF 2 V is 0.1 $\Omega$. It's connected to a resistance of 3.9 $\Omega$. The voltage across the cell will be:
A) 0.5 volt
B) 1.9 volt
C) 1.95 volt
D) 2 volt
To find the voltage across the cell, we utilize the formula for the voltage across an external load when considering the internal resistance of the cell.
The EMF ($E$) of the cell is given as $2$ V, the internal resistance of the cell ($r$) as $0.1 \Omega$, and the external resistance ($R$) as $3.9 \Omega$.
We start by using the voltage division formula $$ V = E - Ir, $$ where $I$ is the current flowing through the circuit. To find $I$, apply Ohm's Law for the total circuit resistance: $$ I = \frac{E}{R + r} = \frac{2}{3.9 + 0.1} = \frac{2}{4} = 0.5 \text{ A}. $$ Then substituting $I$ back into the equation for $V$: $$ V = 2 - 0.5 \times 0.1 = 2 - 0.05 = 1.95 \text{ V}. $$
Thus, the voltage across the cell is 1.95 volts, and the correct option is:
C) 1.95 volt.
Large number of free electrons are present in metals. Why is there no current in the absence of an electric field across it?
In metals, even though there are a large number of free electrons that are constantly in motion, no net electric current is observed in the absence of an external electric field. This is because the individual movements of these electrons are random and isotropic; meaning they move in all directions equally. Consequently, the net motion of electrons in any specific direction cancels out, resulting in no overall current.
When an electric field is applied, however, it influences the electrons to move predominantly in one direction. This directional movement aligns the flow of electrons, thereby generating an electric current. This is why metals conduct electricity when an electric field is present, but exhibit no current in its absence.
"How many electrons will flow through a wire when the charge of $4 \mathrm{C}$ flows for $2 \mathrm{sec}$? Also, find the current."
To find how many electrons flow through a wire when a charge of $4 C$ flows, we divide the total charge by the charge of one electron:
$$ \text{Number of electrons} = \frac{\text{Total charge}}{\text{Charge of one electron}} = \frac{4 , \mathrm{C}}{1.6 \times 10^{-19} , \mathrm{C/electron}} = 2.5 \times 10^{19} \text{ electrons} $$
To find the current flowing in the wire, use the formula for current, which is the rate of flow of charge:
$$ \text{Current} (I) = \frac{\text{Total charge}}{\text{Time period}} = \frac{4 , \mathrm{C}}{2 , \mathrm{s}} = 2 , \mathrm{A} $$
Thus, approximately $2.5 \times 10^{19}$ electrons flow through the wire, and the current in the circuit is $2 , \mathrm{A}$.
If $3.2 \mathrm{~J}$ of work is done in moving $0.4 \mathrm{C}$ of charge in a circuit, find the potential difference (in V) across the circuit. A. 8
Given:
Charge, $ q = 0.4 , \text{C} $
Work done, $ W = 3.2 , \text{J} $
The potential difference $ V $ in volts can be calculated using the formula: $$ V = \frac{W}{q} $$ Substituting the given values: $$ V = \frac{3.2}{0.4} = 8 $$
Therefore, the potential difference across the circuit is $ \mathbf{8 , V} $ (Correct option: A).
The electric current in a charging $R-C$ circuit is given by $i = i_{0} e^{-1 / (R C)}$ where $i_{0}$, $R$, and $C$ are constant parameters of the circuit, and $t$ is time. Find the rate of change of current at: (a) $t = 0$, (b) $t = R C$, (c) $t = 10 R C$.
The solution provided seems to contain some typographical errors and inaccuracies in the interpretation and calculation of the exponential terms. Let's correct and clarify the solution.
Given Function:
The electric current $i$ in the $R-C$ circuit is given by: $$ i = i_0 e^{-t / (R C)} $$ where $i_0$, $R$, $C$ represent the maximum initial current, resistance, and capacitance respectively, and $t$ is the time.
Task:
We need to find the rate of change of the current $\frac{di}{dt}$ at specific values of $t$:
Derivative:
The rate of change of the current with respect to time is obtained by differentiating $i$ with respect to $t$: $$ \frac{di}{dt} = \frac{d}{dt} \left( i_0 e^{-t / (RC)} \right) $$ Using the chain rule of differentiation, we find: $$ \frac{di}{dt} = i_0 \left(-\frac{1}{RC}\right) e^{-t / (RC)} = -\frac{i_0}{RC} e^{-t / (RC)} $$
Solutions:
At $t = 0$:$$ \frac{di}{dt} = -\frac{i_0}{RC} e^{-0 / (RC)} = -\frac{i_0}{RC} e^0 = -\frac{i_0}{RC} $$ The rate of change of current at $t = 0$ is $ \boldsymbol{-\frac{i_0}{RC}}$.
At $t = RC$:$$ \frac{di}{dt} = -\frac{i_0}{RC} e^{-RC / (RC)} = -\frac{i_0}{RC} e^{-1} = -\frac{i_0}{RC e} $$ The rate of change of current at $t = RC$ is $ \boldsymbol{-\frac{i_0}{RC e}}$.
At $t = 10RC$:$$ \frac{di}{dt} = -\frac{i_0}{RC} e^{-10RC / (RC)} = -\frac{i_0}{RC} e^{-10} = -\frac{i_0}{RC e^{10}} $$ The rate of change of current at $t = 10RC$ is $ \boldsymbol{-\frac{i_0}{RC e^{10}}}$.
Each component in these expressions is crucial in identifying how the current's decrease rate accelerates due to the exponential component combined with the circuit parameters.
Eleven $15 \Omega$ and four $7 \Omega$ resistors are connected in series. If a new resistance of $21 \Omega$ is added in series, the change in the effective resistance would be:
A) $21 \Omega$
B) $193 \Omega$
C) $234 \Omega$
D) $15 \Omega"
The correct answer is Option A: $21 \Omega$
The initial configuration involves 11 resistors each of $15 \Omega$ and 4 resistors each of $7 \Omega$ connected in series. As we know, the total resistance for resistors in series is the sum of individual resistances:
$$ R_{\text{eff}} = 11 \times 15 \Omega + 4 \times 7 \Omega $$
Calculating the values gives:
$$ R_{\text{eff}} = 165 \Omega + 28 \Omega = 193 \Omega $$
Adding a new resistor of $21 \Omega$ in series increases the total resistance by the value of the new resistor, which modifies the total effective resistance to:
$$ R_{\text{eff}}' = 193 \Omega + 21 \Omega = 214 \Omega $$
Thus, the change in effective resistance is calculated by subtracting the original resistance from the new total resistance:
$$ \Delta R = R_{\text{eff}}' - R_{\text{eff}} = 214 \Omega - 193 \Omega = 21 \Omega $$
Therefore, the change in effective resistance due to adding the new $21 \Omega$ resistor is $21 \Omega$.
The heat produced in a conductor varies linearly with the current passing through it.
A) True
B) False
The correct answer is Option B: False.
The statement that "the heat produced in a conductor varies linearly with the current passing through it" is incorrect. The heat produced, or energy dissipated, in a conductor due to an electric current is actually proportional to the square of the current. This relationship can be expressed by the formula: $$ H = I^2 R t $$ where:
$H$ is the heat energy,
$I$ is the current through the conductor,
$R$ is the resistance of the conductor,
$t$ is the time for which the current flows.
Thus, the heat depends quadratically, not linearly, on the current.
Why are copper, aluminum, and other metals used for making connecting wires?
A They are good conductors.
B They are good insulators.
C They are cheap and easily available.
D They protect us from electric shock.
The correct answer is A - They are good conductors.
Copper, aluminum, and other metals are widely used for making connecting wires primarily because they are good conductors of electricity. This property allows for efficient transmission of electric current through the wires.
Why does the current always remain the same in a series circuit? When the charges encounter a resistor, wouldn't the speed of flow decrease? If the speed decreases, shouldn't the rate of flow decrease at the same time?
In series circuits, a common question arises about the behavior of current when it passes through a resistor. It's important to clarify that a resistor does not decrease the current; rather, it introduces resistance that causes a voltage drop across the resistor but not a reduction in current itself.
According to Ohm's Law, represented by the equation: $$ V = IR $$ where ( V ) is the voltage, ( I ) is the current, and ( R ) is the resistance, it explains the relationship between voltage, current, and resistance. If you have a constant voltage source in a simple circuit and the resistance is increased, the current through the entire circuit reduces because the voltage remains constant while the resistance increases. Conversely, with a constant current source, the current remains unchanged regardless of changes in the resistance; instead, the voltage drop across the resistor increases.
This brings us to a critical principle in circuits known as Kirchhoff's Current Law, which states that the sum of currents entering a junction must equal the sum of currents exiting the junction. In a series circuit, where resistors are connected one after another without branching, Kirchhoff's law indicates that the current remains constant through every component. Essentially:
The current entering the first resistor $ ( I_1 )$ is the same as the current exiting it and entering the next resistor $( I_2 )$, and so forth across all resistors in the series.
Concisely, in a simple series circuit, the current is the same throughout the circuit because there are no alternate paths (branches) for the current to divert to other components. Hence, even though the resistors impose resistance and cause voltage drops, they do not affect the rate at which charge flows through the circuit, ensuring the current remains consistent along the path.
You are given 48 cells, each of emf $2 \mathrm{~V}$ and internal resistance $1 \mathrm{ohm}$. How will you connect them so that the current through an external resistance of 3 ohms is the maximum?
A) 8 cells in series, 6 such groups in parallel.
B) 12 cells in series, 4 such groups in parallel.
C) 16 cells in series, 3 such groups in parallel.
D) 24 cells in series, 2 such groups in parallel.
Solution Overview:To find the best configuration for connecting 48 cells (each with an electromotive force (emf) of $2 \text{ V}$ and internal resistance of $1 \text{ ohm}$) such that the current through an external resistance of $3 \text{ ohms}$ is maximized, we can evaluate different combinations of series and parallel connections of the cells.
Step-by-Step Solution:
Let $m$ cells be connected in series and $n$ such groups be parallelly connected. If:
The emf of each cell is $E$
The internal resistance per cell is $r$
Total EMF and Resistance Formulas:
When cells are connected in series, the total emf and resistance are: $$ \text{Total EMF} = mE, \quad \text{Total Resistance} = mr $$
Connecting $n$ such groups in parallel, the effective internal resistance becomes: $$ \text{Effective Resistance} = \frac{mr}{n} $$
Total Current, $I$, through the external resistance $R$ can be represented by the formula: $$ I = \frac{mE}{R + \frac{mr}{n}} $$ So fundamentally, maximizing current $I$ relates to minimizing the denominator in the above equation. Let us re-arrange the current expression: $$ I = \frac{mnE}{nR + mr} $$
Optimization Condition:Using the property that the current $I$ will be maximal when the denominator is minimal, we find that for minimal voltage drop across internal resistances: $$ nR = mr $$
Given that $R = 3, \Omega$, $r = 1, \Omega$, and rearranging the above: $$ 3n = m $$ Together with the condition $mn = 48$, we have: $$ \frac{m \times m}{3} = 48 $$ Which simplifies to: $$ m^2 = 144 \quad \text{thus,} \quad m = 12 $$ And consequently: $$ n = \frac{m}{3} = 4 $$
Conclusion:Option B (12 cells in series, 4 such groups in parallel) is the optimal configuration to maximize the current through the external resistance of $3, \Omega$. This configuration maximizes the use of all cells while perfectly balancing the series-parallel arrangement as derived.
A. The current in the reverse-biased condition is generally very small.
B. The current in the reverse-biased condition is small, but the forward-biased current is independent of the bias voltage.
C. The reverse-biased current is strongly dependent on the applied bias voltage.
D. The forward-biased current is very small in comparison to the reverse-biased current.
The correct option is B. The current in the reverse-biased condition is small, but the forward-biased current is independent of the bias voltage.
In reverse bias, a pn junction conducts virtually no current. The application of a reverse bias voltage repels majority carriers away from the junction, effectively expanding the non-conducting depletion region. The only significant current that flows in reverse bias is a temperature-dependent leakage current, which is generally smaller than a microampere ($ \mu A $) in small silicon diodes.
Electric bulb was invented by:
A) Thomas Alva Edison
B) Albert Einstein
C) Isaac Newton
D) Louis Pasteur
The correct answer is A) Thomas Alva Edison.
Thomas Alva Edison invented the first commercially viable electric light bulb in 1879. This invention significantly advanced the use of electric light in homes and industries, paving the way for modern electrical lighting.
In the given setup, resistors A, B, & C are of resistance 1 ohm each. Find the current through resistor A and potential difference across it, if the resistance of the wire is zero.
A) 2 A, 0V
B) 1 A, 2V
C) 4 A, 2V
D) 0A, 0V
The correct option is D) 0A, 0V.
In the given circuit, resistors A, B, and C each have a resistance of 1 ohm. If the resistance of the parallel wire is zero ohms, current will preferentially flow through the path that presents the least resistance. This path is the zero-resistance wire. Thus, no current flows through resistor A, which implies that the current through resistor A is I = 0 A.
We apply Ohm's Law, which is stated as: $$ V = I \times R $$ Substituting the values into the equation, we find: $$ V = 0 , \text{A} \times 1 , \Omega = 0 , \text{V} $$
Therefore, both the current through resistor A and the potential difference across it are zero.
"A resistance of $10 \Omega$ is bent to form a closed circle. What is the effective resistance between the two ends of any diameter of this circuit?"
When the resistance of $10 \Omega$ is bent into a closed circle, the resistance splits evenly along each half if tested across any diameter.
We can consider two resistors of equal resistance, each with a resistance of: $$ \frac{R}{2} = \frac{10 \Omega}{2} = 5 \Omega $$ These two resistors are now in parallel across the diameter. For resistors ( R_1 ) and ( R_2 ) in parallel, the formula for equivalent resistance ($ R_{\text{eq}} $) is given by: $$ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} $$ Substituting ( R_1 = R_2 = 5 \Omega ): $$ \frac{1}{R_{\text{eq}}} = \frac{1}{5 \Omega} + \frac{1}{5 \Omega} = \frac{2}{5 \Omega} $$
Hence, the equivalent resistance $ R_{\text{eq}} $ is: $$ R_{\text{eq}} = \frac{5 \Omega}{2} = 2.5 \Omega $$
Therefore, the effective resistance between the ends of any diameter of this circle is $2.5 \Omega$.
Consider the situation shown in the figure. The wire $PQ$ has mass $m$, resistance $r$ and can slide on the smooth, horizontal parallel rails separated by a distance $l$. The resistance of the rails is negligible. A uniform magnetic field $B$ exists in the rectangular region and a resistance $R$ connects the rails outside the field region. At $t=0$, the wire $PQ$ is pushed towards the right with a speed $v_{0}$. Find:
(a) The current in the loop at an instant when the speed of the wire $PQ$ is $v$. (b) The acceleration of the wire at this instant. (c) The velocity $v$ as a function of $x$. (d) The maximum distance the wire will move.
(a) Current in the loop when the speed of the wire (PQ) is (v)
The electromotive force (emf) induced in the wire is given by: $$ \text{emf} = Blv $$
The total resistance in the circuit is: $$ \text{Resistance} = r + R $$
Therefore, the current in the loop is: $$ \text{Current} = \frac{Blv}{r + R} $$
(b) Acceleration of the wire at this instant
The force acting on the wire due to the magnetic field is given by ( \text{Force} = i l B ): $$ \text{Force} = \frac{BlvlB}{r + R} = \frac{B^2 l^2 v}{r + R} $$
The acceleration of the wire is: $$ a = \frac{ \text{Force} }{ m } = \frac{ B^2 l^2 v }{ m ( r + R ) } $$
(c) Velocity (v) as a function of (x)
The deceleration opposite to the direction of motion is: $$ a = -\frac{B^2 l^2 v}{m (r + R)} $$
Using the differential equation for velocity: $$ \frac{dv}{dx} = -\frac{B^2 l^2 v}{m (r + R)} $$
Integrate both sides: $$ \int_{v_0}^{v} \frac{1}{v} dv = -\int_0^{x} \frac{B^2 l^2}{m(r + R)} dx $$
This gives: $$ \ln \left( \frac{v}{v_0} \right) = -\frac{B^2 l^2}{m(r + R)} x $$
Solving for (v): $$ v = v_0 e^{-\frac{B^2 l^2}{m(r + R)} x} $$
(d) Maximum distance the wire will move
The wire will stop when its velocity becomes zero. Set (v) to zero in the equation for (v): $$ 0 = v_0 e^{-\frac{B^2 l^2}{m(r + R)} x_{\text{max}}} $$
Taking the natural logarithm on both sides results in: $$ \ln \left( \frac{0}{v_0} \right) = -\frac{B^2 l^2}{m(r + R)} x_{\text{max}} $$
As (\ln(0)) approaches (-\infty), ( x_{\text{max}} ) can be directly derived from the integral form: $$ x_{\text{max}} = \frac{m (r + R) v_0}{B^2 l^2} $$
If one of the two wires connected to an electric bulb is called 'Live wire', then what is the other one called?
Option 1) Neutral wire
Option 2) Fuse wire
Option 3) Bulb wire
Option 4) Flow wire
The correct option is 1) Neutral wire.
The two electrical supply wires that connect to the terminals of an electric bulb are called 'Live' wire and 'Neutral' wire.
Find the equivalent resistance across the points $A$ and $B$ in the given circuit.
A) $1 \Omega$ B) $2 \Omega$ C) $4 \Omega$ D) $8 \Omega$
The correct option is B) $2 \Omega$.
Here's the detailed step-by-step explanation:
Series Combination of Resistors 'a' and 'b':
The resistors labeled '$a$' and '$b$' are connected in series.
The equivalent resistance, denoted as $R_1$, can be calculated as: $$ R_1 = 2\ \Omega + 2\ \Omega = 4\ \Omega $$
Series Combination of Resistors 'c' and 'd':
Similarly, the resistors labeled '$c$' and '$d$' are also connected in series.
The equivalent resistance, denoted as $R_2$, is: $$ R_2 = 2\ \Omega + 2\ \Omega = 4\ \Omega $$
Parallel Combination of $R_1$ and $R_2$:
Now, $R_1$ and $R_2$ are in parallel. To find the equivalent resistance for resistors in parallel, we use the formula: $$ \frac{1}{R_{\text{equ}}} = \frac{1}{R_1} + \frac{1}{R_2} $$
Substituting the values of $R_1$ and $R_2$: $$ \frac{1}{R_{\text{equ}}} = \frac{1}{4\ \Omega} + \frac{1}{4\ \Omega} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} $$
Solving for $R_{\text{equ}}$:
Finally, the equivalent resistance $R_{\text{equ}}$ is: $$ R_{\text{equ}} = 2\ \Omega $$
Therefore, the equivalent resistance across points $A$ and $B$ is $2 \Omega.
The resistance $R$ of a wire is found by determining its length $l$ and radius $r$. The percentage errors in measurement of $l$ and $r$ are respectively $1%$ and $2%$. The percentage error in measurement of $R$ is:
A. 4 %
B. 5 %
C. 3 %
D. 1 %
To determine the percentage error in the resistance $R$ of a wire, given the percentage errors in the measurements of its length $l$ and radius $r$, we can use the dependency of resistance on these quantities.
Step-by-Step :
Resistance Formula:The resistance $R$ of a wire can be expressed as: $$ R = \rho \frac{l}{A} $$ where $\rho$ is the resistivity of the material, $l$ is the length of the wire, and $A$ is the cross-sectional area. For a wire with a circular cross-section, the area $A$ is given by: $$ A = \pi r^2 $$ Thus, the resistance formula can be rewritten as: $$ R = \rho \frac{l}{\pi r^2} $$
Identifying Variables:Here, $\rho$ and $\pi$ are constants. Therefore, the resistance $R$ depends only on the length $l$ and the radius $r$.
Percentage Error Formula:The overall percentage error in $R$ is found by considering the percentage errors in $l$ and $r$. For a function $R = \frac{l}{r^2}$, the percentage error in $R$ ($\Delta R$) can be approximated by: $$ \frac{\Delta R}{R} \times 100 \approx \left(\frac{\Delta l}{l} \times 100\right) + 2 \left(\frac{\Delta r}{r} \times 100\right) $$
Given Percentage Errors:
The percentage error in length $l$ is $1%$.
The percentage error in radius $r$ is $2%$.
Calculating the Percentage Error in $R$:Substituting the given values into the error formula: $$ \text{Percentage error in } R \approx 1% + 2 \times 2% $$ Performing the calculation: $$ = 1% + 4% = 5% $$
Conclusion:
The percentage error in the measurement of the resistance $R$ is $5%$.
Final Answer: $\boxed{5\%}$
Current density in a cylindrical wire of radius $R$ is given as
$$J=({ \begin{array}{l} J_{0}\left(\frac{x}{R}-1\right) \text{ for } 0 \leq x < \frac{R}{2} \ J_{0} \frac{x}{R} \text{ for } \frac{R}{2} \leq x \leq R \end{array} })$$
The current flowing in the wire is
A. $\frac{7}{24} \pi J_{0} R^{2}$
B. $\frac{1}{6} \pi J_{0} R^{2}$
C. $\frac{7}{12} \pi J_{0} R^{2}$
D. $\frac{5}{12} \pi J_{0} R^{2}$
The correct option is D: $\frac{5}{12} \pi J_{0} R^{2}$
Given:
$$J=({ \begin{array}{l} J_{0}\left(\frac{x}{R}-1\right) \text{ for } 0 \leq x < \frac{R}{2} \ J_{0} \frac{x}{R} \text{ for } \frac{R}{2} \leq x \leq R \end{array} })$$
Let us consider a cylindrical wire with differential thickness $\mathrm{dx}$ and the area of the element to be $dA = 2 \pi x , dx$.
Applying the concept of current density:
$$ \mathrm{I}=\int \overrightarrow{\mathrm{J}} \cdot \overrightarrow{\mathrm{dA}} $$
Since $\vec{J}$ and $\overrightarrow{d A}$ are parallel:
$$ I=\int_{0}^{R} J \cdot dA = \int_{0}^{R} J \cdot 2\pi x , dx $$
We can split this integral into two parts corresponding to the given piecewise function for $J$:
$$ I=\int_{0}^{R/2} J_0 \left(\frac{x}{R} - 1\right) 2 \pi x , dx + \int_{R/2}^{R} J_0 \frac{x}{R} 2 \pi x , dx $$
Evaluating these integrals:
First part:
$$ I_1 = 2 \pi J_0 \int_{0}^{R/2} \left(\frac{x^2}{R} - x\right) dx $$
$$ I_1 = 2 \pi J_0 \left[\frac{x^3}{3R} - \frac{x^2}{2}\right]_{0}^{R/2} $$
$$ I_1 = 2 \pi J_0 \left(\frac{\left(\frac{R}{2}\right)^3}{3R} - \frac{\left(\frac{R}{2}\right)^2}{2}\right) $$
$$ I_1 = 2 \pi J_0 \left(\frac{R^3}{24R} - \frac{R^2}{8}\right) $$
$$ I_1 = 2 \pi J_0 \left(\frac{R^2}{24} - \frac{R^2}{8}\right) $$
$$ I_1 = 2 \pi J_0 \left(\frac{R^2 - 3R^2}{24}\right) $$
$$ I_1 = 2 \pi J_0 \left(\frac{-2R^2}{24}\right) $$
$$ I_1 = \frac{-\pi J_0 R^2}{6} $$
Second part:
$$ I_2 = 2 \pi J_0 \int_{R/2}^{R} \frac{x^2}{R} , dx $$
$$ I_2 = 2 \pi J_0 \frac{1}{R} \int_{R/2}^{R} x^2 , dx $$
$$ I_2 = 2 \pi J_0 \frac{1}{R} \left[\frac{x^3}{3}\right]_{R/2}^{R} $$
$$ I_2 = 2 \pi J_0 \frac{1}{R} \left(\frac{R^3}{3} - \frac{\left(\frac{R}{2}\right)^3}{3}\right) $$
$$ I_2 = 2 \pi J_0 \frac{1}{R} \left(\frac{R^3}{3} - \frac{R^3}{24}\right) $$
$$ I_2 = 2 \pi J_0 \frac{1}{R} \left(\frac{8R^3 - R^3}{24}\right) $$
$$ I_2 = 2 \pi J_0 \frac{7R^3}{24R} $$
$$ I_2 = \frac{7 \pi J_0 R^2}{12} $$
Summing both parts:
$$ I = I_1 + I_2 = \frac{-\pi J_0 R^2}{6} + \frac{7 \pi J_0 R^2}{12} $$
Finding the common denominator:
$$ I = \frac{-2 \pi J_0 R^2}{12} + \frac{7 \pi J_0 R^2}{12} $$
$$ I = \frac{5 \pi J_0 R^2}{12} $$
Thus, the correct answer is D: $\frac{5}{12} \pi J_{0} R^{2}$
If electrons are not used up in electrical appliances, what actually happens so that the appliance starts working?
When electrons are forced to move in synchronicity, they can produce heat and — even more impressively — turn the wires they travel through into magnets. This heat can boil water and cause light bulbs to glow, while the magnets can cause objects to move. These two principles are the foundation of the 'magic' behind every electrical appliance.
Wires are made of metal, which contains loosely bound electrons buzzing around. When these electrons move in an organized manner, you get an electric current flowing through the wire. All wires heat up slightly when a current runs through them because the moving electrons collide with the metal atoms. Each collision results in energy being released in the form of heat.
Every appliance with moving parts contains an electric motor. These motors are designed to spin whenever power is supplied. As a result, any parts attached to the motor, such as fan blades, wheels, or washing machine tubs, also spin. This spinning occurs only when current is flowing — when electrons in the wire are organized into a current. Moving electrons make a motor spin by turning wires into magnets, which are exceptional at inducing motion.
Draw a circuit diagram to show how two 4V electric lamps can be lit brightly from two 2V cells.
To light two 4V electric lamps brightly using two 2V cells, follow these steps:
Connect the two 2V cells in series. When you connect cells in series, their voltages add up. Therefore, two 2V cells in series will produce a total voltage of 4V.
Connect the two 4V lamps in parallel with the series-connected cells. Connecting the lamps in parallel ensures each lamp receives the full 4V from the combined cells, allowing them to light up brightly.
Here is the circuit diagram illustrating the setup:
Explanation of the setup:
Series Connection of Cells: The positive terminal of the first cell is connected to the negative terminal of the second cell.
Parallel Connection of Lamps: Each lamp is connected across the same two points of the circuit, ensuring each one receives the same voltage from the cells.
By arranging the cells and lamps this way, you ensure that each lamp receives the necessary 4V to operate brightly.
The volume of a liquid (V) flowing per second through a cylindrical tube depends upon the pressure gradient (p/l), radius of the tube (r), and coefficient of viscosity (η) of the liquid. By dimensional analysis, the correct formula is:
V ∝ P r^4 / η l
Therefore, the correct formula is:
A) $ V \propto \frac{P r^4}{\eta l} $
B) $ V \propto \frac{P r}{\eta l} $
C) $ V \propto \frac{P l}{\eta r} $
D) None
To determine the rate of flow of a liquid, $V$, in a cylindrical tube, considering it depends on the pressure gradient $ \left( \frac{P}{l} \right) $, the radius of the tube $ r $, and the coefficient of viscosity $ \eta $ of the liquid, we can use dimensional analysis to derive the correct formula.
Dimensional Analysis
Dimensions of Volume Flow Rate ($ V $):
The volume of liquid flowing per second (( V )) is essentially a rate of flow or flux. Its dimensions are: $$ [V] = [L^3 T^{-1}] $$
Dimensions of Radius ($ r $):
Radius is a length measure, so: $$ [r] = [L] $$
Dimensions of Pressure Gradient $ \left( \frac{P}{l} \right) $:
The pressure gradient combines pressure $ (P) $ and length $ (L) $. Pressure is force per unit area which is $ M L^{-1} T^{-2} $. Thus, the dimensions become: $$ \left[ \frac{P}{l} \right] = [M L^{-2} T^{-2}] $$
Dimensions of Coefficient of Viscosity ($ \eta $):
Viscosity has dimensions: $$ [\eta] = [M L^{-1} T^{-1}] $$
Assuming: $$ V \propto r^a \left( \frac{P}{l} \right)^b \eta^c $$
Equating Dimensional Powers
Substitute the dimensions: $$ [L^3 T^{-1}] \propto [L]^a [M L^{-2} T^{-2}]^b [M L^{-1} T^{-1}]^c $$
Combine and simplify: $$ [L^3 T^{-1}] \propto [L^a] [M^b L^{-2b} T^{-2b}] [M^c L^{-c} T^{-c}] $$
$$ [L^3 T^{-1}] \propto [L^{a - 2b - c} M^{b + c} T^{-2b - c}] $$
Now, match the dimensions from both sides:
For Length ($ L $):$$ a - 2b - c = 3 $$
For Mass ($ M $):$$ b + c = 0 $$
For Time ($ T $):$$ -2b - c = -1 $$
Solving the System of Equations
From $ b + c = 0 $, we have $ c = -b $.
Substitute $ c = -b $ into the equations:
$ a - 2b - (-b) = 3 $ $$ a - b = 3 \quad \text{(i)} $$
$ -2b - (-b) = -1 $ $$ -b = -1 \quad \Rightarrow \quad b = 1 \quad \text{(ii)} $$
From (ii), $ b = 1 $. Substitute $ b = 1 $ into $ b + c = 0 $: $$ 1 + c = 0 \quad \Rightarrow \quad c = -1 $$
Now, use $ b = 1 $ in $ a - b = 3 $: $$ a - 1 = 3 \quad \Rightarrow \quad a = 4 $$
Final Formula
Compressing all results: $$ V \propto r^4 \left( \frac{P}{l} \right)^1 \eta^{-1} $$
Thus, the correct formula for the rate of flow of liquid is: $$ V \propto \frac{P \cdot r^4}{\eta \cdot l} $$
Conclusion
The correct formula is: [ \boxed{V \propto \frac{P \cdot r^4}{\eta \cdot l}} ]
So, the correct option is: A) $ V \propto \frac{P r^4}{\eta l} $
A current balance (or ampere balance) is a device for measuring currents. The current to be measured is arranged to go through two long parallel wires of equal length in opposite directions, one of which is linked to the pivot of the balance. The resulting repulsive force on the wire is balanced by putting a suitable mass in the scale pan hanging on the other side of the pivot. In one measurement, the mass in the scale pan is 30.0 g, the length of the wires is 50.0 cm each, and the separation between them is 10.0 mm. What is the value of the current being measured? Take g = 9.80 m/s^2 and assume that the arms of the balance are equal.
To determine the current (I) using a current balance, we're given the following data:
Mass in the scale pan, $m = 30.0$ g
Length of the wires, $l = 50.0$ cm
Separation between the wires, $d = 10.0$ mm
Acceleration due to gravity, $g = 9.80$ m/s(^2)
Conversion of Units
Mass ( m ) to kg: $$ m = 30.0 , \mathrm{g} = 30.0 \times 10^{-3} , \mathrm{kg} $$
Length ( l ): $$ l = 50.0 , \mathrm{cm} = 0.50 , \mathrm{m} $$
Separation ( d ): $$ d = 10.0 , \mathrm{mm} = 10.0 \times 10^{-3} , \mathrm{m} = 0.01 , \mathrm{m} $$
Calculation of the Force (F)
The force due to the mass is: $$ F = m \cdot g = 30.0 \times 10^{-3} , \mathrm{kg} \times 9.80 , \mathrm{m/s^2} $$
Simplifying, $$ F = 0.30 , \mathrm{kg} \times 9.80 , \mathrm{m/s^2} = 0.294 , \mathrm{N} $$
Magnetic Force Between the Wires
The magnetic force (F) between two parallel currents is given by: $$ F = \frac{\mu_0}{2\pi} \cdot \frac{I^2 l}{d} $$
Given that $$ \mu_0 = 4\pi \times 10^{-7} , \mathrm{T \cdot m/A} $$
Substitute the values: $$ F = 10^{-7} \cdot \frac{2 I^2}{0.01} \cdot 0.5 $$
$$ F = 10^{-5} \cdot I^2 $$
Equating the Forces
The force calculated from the mass should be equal to the magnetic force: $$ 0.294 , \mathrm{N} = 10^{-5} I^2 $$
Solving for ( I ): $$ I^2 = \frac{0.294 , \mathrm{N}}{10^{-5}} $$ $$ I^2 = 29400 $$ $$ I = \sqrt{29400} = 171.5 , \mathrm{A} $$
Final Answer: The current being measured is 171.5 A.
A wire of length $L$ and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by $\Delta T$ in time $t. N$ number of similar cells is now connected in series with a wire of the same material and cross section but of length $2 \mathrm{~L}$. The temperature of the wire is raised by the same amount $\Delta T$ in the same time $t$. The value of $N$ is:
To determine the value of ( N ), we can compare the conditions given in the problem. Here, the goal is to ensure that the wire's temperature is raised by (\Delta T) in the same time ( t ), using identical cells with negligible internal resistance.
Given:
Length of the original wire: ( L )
Number of cells initially: 3
The wire's heating condition: (\Delta T) in time ( t )
We need:
Length of the new wire: ( 2L )
Same temperature rise ( \Delta T ) in the same time ( t )
Number of cells required: ( N )
Step-by-Step
By the relationship given, we know: [ I^2 R \Delta t = \text{Thermal energy responsible for temperature increase to } \Delta T ]
For the second scenario with a wire of length ( 2L ): [ I^{\prime 2} R' \Delta t = 2 I^2 R \Delta t ]
Since resistance is proportional to length, for a wire ( 2L ): ( R' = 2R ). Hence, [ I^{\prime 2} (2R) \Delta t = 2 I^2 R \Delta t ]
This implies: [ I^{\prime 2} 2R = 2 I^2 R ] [ I^{\prime} = I ]
The current ( I ) in the first case with 3 cells: [ I = \frac{3E}{R} ]
In the second case with ( N ) cells: [ I^{\prime} = \frac{N E}{2R} ]
Given ( I^{\prime} = I ): [ \frac{N E}{2R} = \frac{3 E}{R} ]
Solving for ( N ): [ N = 6 ]
Thus, the value of ( N ) is 6.
💡 Have more questions?
Ask Chatterbot AINCERT Solutions - Current Electricity | NCERT | Physics | Class 12
The storage battery of a car has an emf of $12 \mathrm{~V}$. If the internal resistance of the battery is $0.4 \Omega$, what is the maximum current that can be drawn from the battery?
The maximum current that can be drawn from the battery is
$$ \mathbf{30 , \text{A}} $$
A battery of emf $10 \mathrm{~V}$ and internal resistance $3 \Omega$ is connected to a resistor. If the current in the circuit is $0.5 \mathrm{~A}$, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Results
Resistance of the Resistor:$$ R_{\text{resistor}} = \frac{8.5 \mathrm{~V}}{0.5 \mathrm{~A}} = 17 \Omega $$
Terminal Voltage of the Battery:$$ V = \varepsilon - V_{\text{internal}} = 10 \mathrm{~V} - 1.5 \mathrm{~V} = 8.5 \mathrm{~V} $$
Summary:
Resistance of the Resistor: $ \mathbf{17 \Omega} $
Terminal Voltage of the Battery: $ \mathbf{8.5 \mathrm{~V}} $
These values indicate that the resistor has a resistance of 17 ohms and the terminal voltage of the battery when the circuit is closed is 8.5 volts.
At room temperature $\left(27.0^{\circ} \mathrm{C}\right)$ the resistance of a heating element is $100 \Omega$. What is the temperature of the element if the resistance is found to be $117 \Omega$, given that the temperature coefficient of the material of the resistor is $1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$.
The temperature of the heating element when its resistance is (117 \Omega) is 1027°C.
A negligibly small current is passed through a wire of length $15 \mathrm{~m}$ and uniform cross-section $6.0 \times 10^{-7} \mathrm{~m}^{2}$, and its resistance is measured to be $5.0 \Omega$. What is the resistivity of the material at the temperature of the experiment?
The resistivity of the material is: $$ \rho = 2 \times 10^{-7} , \Omega \cdot \text{m} $$
So, the resistivity at the temperature of the experiment is $ \boldsymbol{2 \times 10^{-7} , \Omega \cdot \text{m}} $.
A silver wire has a resistance of $2.1 \Omega$ at $27.5^{\circ} \mathrm{C}$, and a resistance of $2.7 \Omega$ at $100{ }^{\circ} \mathrm{C}$. Determine the temperature coefficient of resistivity of silver.
The temperature coefficient of resistivity (\alpha) for silver is approximately:
$$ \alpha \approx 0.00394 , \mathrm{^{\circ}C^{-1}} $$
This value indicates that the resistivity of silver increases by about 0.394% per degree Celsius rise in temperature.
A heating element using nichrome connected to a $230 \mathrm{~V}$ supply draws an initial current of 3.2 A which settles after a few seconds to a steady value of $2.8 \mathrm{~A}$. What is the steady temperature of the heating element if the room temperature is $27.0^{\circ} \mathrm{C}$ ? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is $1.70 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$.
From the calculation:
[ T \approx 867.35 , ^{\circ}\text{C} ]
So, the steady temperature of the heating element is approximately 867.35°C.
A storage battery of emf 8.0 V and internal resistance $0.5 \Omega$ is being charged by a $120 \mathrm{~V}$ dc supply using a series resistor of $15.5 \Omega$. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?
The terminal voltage of the battery during charging is: $$ \mathbf{11.8 \text{ V}} $$
Purpose of Having a Series Resistor in the Charging Circuit
The series resistor in the charging circuit serves several important purposes:
Current Limiting: The series resistor limits the charging current to a safe value, preventing excessive current that could damage the battery or cause overheating.
Voltage Regulation: By dropping some voltage across the resistor, it helps to maintain a more stable and controlled charging voltage at the battery terminals.
Safety: It acts as a protective component to safeguard against potential short circuits and overcurrent conditions.
Thus, the series resistor is crucial for ensuring safe and efficient battery charging.
The number density of free electrons in a copper conductor estimated in Example 3.1 is $8.5 \times 10^{28} \mathrm{~m}^{-3}$. How long does an electron take to drift from one end of a wire $3.0 \mathrm{~m}$ long to its other end? The area of cross-section of the wire is $2.0 \times 10^{-6} \mathrm{~m}^{2}$ and it is carrying a current of 3.0 A.
First, we computed the denominator:
$$ 8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.0 \times 10^{-6} = 27200 $$
Next, we compute the drift velocity:
$$ v_d = \frac{3}{27200} \approx 0.0001103 , \text{m/s} $$
Finally, the time taken ( t ) to drift the length of the wire:
$$ t = \frac{L}{v_d} = \frac{3.0}{0.0001103} \approx 27200 , \text{s} $$
Therefore, an electron takes approximately 27200 seconds to drift from one end of the $3.0 , \text{m}$ wire to the other. This is equal to about 7.56 hours.
💡 Have more questions?
Ask Chatterbot AINotes - Current Electricity | Class 12 NCERT | Physics
Comprehensive Current Electricity Class 12 Notes: Understanding Basic Concepts
Introduction to Current Electricity
Understanding current electricity forms the bedrock of classical and modern physics. It refers to the flow of electric charges through a conductor and underpins the operation of numerous electrical devices and technologies.
Electric Current
Definition and Formula
Electric current (I) is defined as the net charge (Q) flowing through a cross-sectional area of a conductor per unit time (t). It is expressed mathematically as: [ I = \frac{Q}{t} ]
Units of Measurement (Amperes)
The SI unit of electric current is the ampere (A), named after French physicist André-Marie Ampère.
Steady vs. Non-Steady Currents
- Steady Current: Constant flow of charges.
- Non-Steady Current: Varying flow of charges with time.
Electric Currents in Conductors
Effect of Electric Field on Electrons
When an electric field is applied to a conductor, free electrons experience a force causing them to move, thereby generating a current.
Concept of Drift Velocity
The average velocity of electrons in a conductor due to an electric field is termed drift velocity.
Ohm's Law
Definition and Mathematical Expression
Ohm's Law states that the current (I) flowing through a conductor is directly proportional to the voltage (V) across its ends, provided the temperature remains constant. It's formulated as: [ V = IR ] where R is the resistance of the conductor.
Application in Circuits
Ohm's Law is used to calculate the resistance, voltage, and current in series and parallel circuits.
Derivation and Units of Resistance
Resistance (R) is expressed in ohms (Ω). One ohm is the resistance when a voltage of one volt produces a current of one ampere.
Resistance and Resistivity
Definition and Differences
- Resistance (R): Opposition to the flow of current in a conductor.
- Resistivity (ρ): Material property indicating how strongly a material opposes the flow of electric current.
Factors Affecting Resistance
- Length (l) of the conductor
- Cross-sectional area (A)
- Material of the conductor
Temperature Dependence of Resistivity
Resistivity typically increases with temperature in conductors.
Drift of Electrons and Origin of Resistivity
Collisions and Drift Velocity
Electrons collide with atoms in a conductor, affecting their drift velocity. The resistivity (ρ) of a material is given by: [ \rho = \frac{m}{n e^2 \tau} ] where m is electron mass, n is charge carrier density, e is electron charge, and τ is the relaxation time.
Calculation of Drift Speed
Drift speed (v_d) is calculated as: [ v_d = \frac{I}{n e A} ]
Limitations of Ohm's Law
Ohm's Law may not hold under certain conditions:
- Non-linear V-I relationship
- Dependence on the sign of voltage
- Non-unique V-I relationship
Electrical Power and Energy
Formulas for Power Calculation
- Power (P) dissipated in a resistor is given by: [ P = IV ] Using Ohm's Law, it can also be expressed as: [ P = I^2 R ] or [ P = \frac{V^2}{R} ]
Practical Examples and Problems
Example: A resistor with a resistance of 5 Ω has a current of 2 A flowing through it. The power dissipated is: [ P = I^2 R = 2^2 \times 5 = 20 \text{ W} ]
Cells and Electromotive Force (emf)
Emf and Internal Resistance
A cell's emf (ε) is the potential difference when no current is drawn. The internal resistance (r) of a cell affects the terminal voltage (V), given by: [ V = \varepsilon - I r ]
Series and Parallel Combination of Cells
- Series: Emfs add up, total internal resistance is the sum of individual resistances.
- Parallel: Effective emf is a weighted average based on internal resistances.
Kirchhoff's Rules
Junction Rule
The sum of currents entering a junction equals the sum leaving it.
Loop Rule
The algebraic sum of changes in potential around any closed loop is zero.
graph LR
A --> B
B --> C
C --> D
D --> A
E --> B
E --> D
Wheatstone Bridge
Principle and Balance Condition
A Wheatstone Bridge consists of four resistances arranged in a diamond shape. It is balanced when: [ \frac{R_1}{R_2} = \frac{R_3}{R_4} ]
Practical Applications
Used for precise measurement of unknown resistance values.
Mobility and Current Density
Definition of Mobility
Mobility (μ) is the drift velocity per unit electric field: [ \mu = \frac{v_d}{E} ]
Significance of Current Density
Current density (j) is the electric current per unit area: [ j = \frac{I}{A} ]
Electrical Properties of Different Materials
Conductors, Semiconductors, and Insulators
- Conductors: Low resistivity (e.g., metals)
- Semiconductors: Intermediate resistivity, decreases with temperature
- Insulators: High resistivity (e.g., glass, rubber)
Role of Temperature and Impurities
Temperature and impurities significantly affect the resistivity of materials.
Important Equations and Units
Summary of Key Formulas
- Ohm's Law: [ V = IR ]
- Power: [ P = IV = I^2R = \frac{V^2}{R} ]
- Resistivity: [ \rho = \frac{R A}{l} ]
Units of Measurement
- Current: Amperes (A)
- Voltage: Volts (V)
- Resistance: Ohms (Ω)
- Power: Watts (W)
Conclusion
Current electricity is a fundamental concept in physics, essential for understanding and designing electrical circuits and devices. A clear grasp of the principles such as Ohm's Law, resistivity, and Kirchhoff's Rules is crucial for students at the Class 12 level.
🚀 Learn more about Notes with Chatterbot AI