Electricity - Class 10 Science - Chapter 11 - Notes, NCERT Solutions & Extra Questions
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Notes - Electricity | Class 10 NCERT | Science
Electricity Class 10 Notes: Detailed Concepts and Important Points
Electricity holds a pivotal role in our modern society, powering our homes, industries, hospitals, and various other sectors. It is crucial for students to grasp the fundamental concepts of electricity, especially those in Class 10. This comprehensive guide will help you understand the key topics and provide clear, concise notes for your studies.
Introduction to Electricity
Electricity is a form of energy resulting from the existence of charged particles such as electrons or protons. It has become an indispensable part of modern life, offering a convenient and controllable source of power for various applications.
Understanding Electric Current and Circuit
What is an Electric Circuit?
An electric circuit is a continuous and closed path through which electric current flows. It comprises components such as cells (batteries), bulbs, ammeters, and switches.
Flow of Electric Current in Conductors
In conductors like metallic wires, electric current is the flow of electric charges. This flow can be visualized as electrons moving through the wire, creating a current.
Measuring Electric Current
Units of Electric Current
Electric current is measured in amperes (A), named after André-Marie Ampère, a French scientist. Smaller units include milliamperes (mA) and microamperes (µA).
Instruments Used
An ammeter is an instrument used to measure electric current and is always connected in series with the circuit.
Electric Potential and Potential Difference
Definition and Analytical Explanation
Electric potential difference, also known as voltage, is the work done to move a unit charge from one point to another in a circuit. It’s measured in volts (V).
Units and Measurement Tools
A voltmeter is used to measure the potential difference and is always connected in parallel with the circuit component.
Fundamentals of Ohm's Law
Statement and Formula
Ohm’s Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, provided the temperature remains constant. The formula is V = IR, where V is voltage, I is current, and R is resistance.
Graphical Representation (V-I Graph)
The V-I graph for a resistor is a straight line, indicating a linear relationship between voltage and current.
Practical Applications and Calculations
Ohm's Law is used to calculate the resistance, current, or voltage in an electric circuit.
Resistance and Resistivity
Factors Affecting Resistance
Resistance of a conductor depends on its length, cross-sectional area, and the material it is made of. Longer lengths and smaller cross-sectional areas increase resistance.
Calculating Resistivity
Resistivity is a measure of a material’s ability to resist the flow of electric current, calculated using the formula (\rho = RA/l), where R is resistance, A is cross-sectional area, and l is length.
Difference Between Conductors, Resistors, and Insulators
Conductors allow easy flow of electric current, resistors impede this flow to some extent, and insulators block it completely.
Series and Parallel Circuits
Characteristics of Series Circuits
In a series circuit, components are connected end-to-end, so the same current flows through each component. The total resistance is the sum of all resistances.
Characteristics of Parallel Circuits
In a parallel circuit, components are connected across common points, so the voltage remains the same across each component while the currents can differ. The total resistance is found using $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}$.
Heating Effect of Electric Current
Joule’s Law of Heating
The heat produced in a resistor is given by $H = I^2Rt$, where I is current, R is resistance, and t is time.
Practical Appliances Using Heating Effect
Devices like electric heaters, toasters, and irons utilize the heating effect of electric current to operate.
Electric Power and Energy
Definitions and Units
Electric power is the rate at which electric energy is consumed or dissipated. It is given by (P = VI) and measured in watts (W). Electric energy is the total power consumed over time, measured in watt-hours (Wh) or kilowatt-hours (kWh).
Calculating Power in Electrical Devices
Power can also be calculated using $P = I^2R$ or $P = \frac{V^2}{R}$.
Schematic Diagrams in Electric Circuits
Symbols and Representation
Schematic diagrams use standardized symbols to represent different components, making it easier to visualize and understand circuits.
Drawing and Interpreting Circuit Diagrams
Learning to draw and interpret these diagrams is crucial for solving circuit-related problems.
Problem Solving in Electric Circuits
Use the principles and formulas discussed to solve numerical problems related to electric circuits. Practice with various examples to strengthen your understanding.
Conclusion
Understanding the basics of electricity, including circuits, current, voltage, resistance, and power, is essential for both academic success and practical applications. These Class 10 notes aim to make these concepts clear and accessible.
Additional Resources
Further Reading: Educational books and online articles
Practice Questions: Available in textbooks and online quizzes
Interactive Simulations: Websites providing virtual labs for experimenting with circuits
By following these notes and practicing regularly, you will be well-prepared to tackle any electricity-related topics in your Class 10 exams.
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Extra Questions - Electricity | NCERT | Science | Class 10
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A piece of wire of resistance $R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is $R^{\prime}$, then the ratio $R / R^{\prime}$ is -
(a) $1 / 25$
(b) $1 / 5$
(c) 5
(d) 25
When a wire of resistance $R$ is cut into five equal parts, the length of each part becomes $1/5$ of the original length. Since the resistance $R$ of a wire is directly proportional to its length $(l)$, given by the formula $R = \rho \cdot (l/A)$ where $\rho$ is the resistivity of the material and (A) is the cross-sectional area, each part will have a resistance of (R/5) because the resistivity and cross-sectional area remain unchanged.
When these parts are connected in parallel, the equivalent resistance $R'$ can be found using the formula for parallel resistances:
$$\frac{1}{R'} = \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5}$$
$$\frac{1}{R'} = 5 \cdot \frac{1}{R/5}$$
$$\frac{1}{R'} = \frac{5}{R/5}$$
$$\frac{1}{R'} = \frac{25}{R}$$
Therefore,
$$R' = \frac{R}{25}$$
The ratio $R / R'$ can then be calculated as follows:
$$R / R' = R / \left( \frac{R}{25} \right) = 25$$
Thus, the correct answer is (d) 25.
Which of the following terms does not represent electrical power in a circuit?
(a) $I^{2} R$
(b) $I R^{2}$
(c) $V I$
(d) $V^{2} / R$
Electrical power in a circuit can be represented using various expressions depending on the known quantities (voltage, current, resistance). These expressions are derived from Ohm's law $(V = IR)$, where $V$ represents voltage, (I) represents current, and $R$ represents resistance. The formulas for power $(P)$ are:
$P = I^2R$: Power is directly proportional to the square of the current through the resistor times the resistance.
$P = VI$: Power is the product of voltage across the resistor and the current through it.
$P = \frac{V^2}{R}$: Power is directly proportional to the square of the voltage across the resistor divided by the resistance.
The option that does not represent electrical power in a circuit is:
(b) $IR^2$
This expression is incorrect for computing power in an electrical circuit and does not conform to any of the standard formulas derived from Ohm's law.
An electric bulb is rated $220 \mathrm{~V}$ and $100 \mathrm{~W}$. When it is operated on $110 \mathrm{~V}$, the power consumed will be -
(a) $100 \mathrm{~W}$
(b) $75 \mathrm{~W}$
(c) $50 \mathrm{~W}$
(d) $25 \mathrm{~W}$
The power $ P $ consumed by an electric bulb (or any resistive device) is given by the equation:
$$ P = \frac{V^2}{R} $$
where $V$ is the voltage across the bulb, and $ R $ is the resistance of the bulb.
Given a bulb rated at $220 , \text{V}$ and $ 100 , \text{W}$, we can find its resistance using the rated conditions:
$$ R = \frac{V_{\text{rated}}^2}{P_{\text{rated}}} = \frac{(220)^2}{100}$$
When the bulb is operated at $110 , \text{V} $, the power consumed can be calculated using:
$P' = \frac{V_{\text{new}}^2}{R} = \frac{(110)^2}{R}$
Substituting the value of $R$ from the first equation:
$$ P' = \frac{(110)^2}{\frac{(220)^2}{100}} = \frac{(110)^2 \cdot 100}{(220)^2} $$
Let's compute this value. The power consumed by the electric bulb when operated at $110 , \text{V}$ is $25 , \text{W}$.
Hence, the correct answer is (d) $25 \mathrm{~W}$.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be -
(a) $1: 2$
(b) $2: 1$
(c) $1: 4$
(d) 4:1
To solve this problem, you should first recall how the resistance of a conductor depends on its length and cross-sectional area, and how the heat produced in a resistor (when current passes through it) is given by Joule's law. The resistance (R) of a conductor is given by:
$$ R = \rho \frac{L}{A} $$
where $\rho$ is the resistivity of the material (which is constant for the same material), $L$ is the length, and $A$ is the cross-sectional area of the conductor.
When two resistors are connected in series, their equivalent resistance $(R_{eq_s})$ is the sum of their resistances:
$$ R_{eq_s} = R_1 + R_2 $$
And when connected in parallel, their equivalent resistance ((R_{eq_p})) is given by:
$$ \frac{1}{R_{eq_p}} = \frac{1}{R_1} + \frac{1}{R_2} $$
Since both wires are of equal length and diameter, they have the same resistance ((R)), so for series connection:
$$ R_{eq_s} = R + R = 2R $$
And for parallel connection:
$$ \frac{1}{R_{eq_p}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \Rightarrow R_{eq_p} = \frac{R}{2} $$
The heat produced ((H)) in a given time period in a resistor through which a current (I) flows is given by Joule's law, where:
$$ H = I^2 R t $$
With a constant potential difference ((V)) across the resistors, the currents in the series and parallel cases $I_s$ and $I_p$, can be found using Ohm's law $(V = IR)$, so:
For series: $V = I_s \cdot 2R \Rightarrow I_s = \frac{V}{2R}$
For parallel: $V = I_p \cdot \frac{R}{2} \Rightarrow I_p = \frac{2V}{R}$
Substituting $I_s$ and $I_p$ into Joule's law, we find:
For series: $H_s = (\frac{V}{2R})^2 \cdot 2R \cdot t = \frac{V^2}{2R} \cdot t$
For parallel: $H_p = (\frac{2V}{R})^2 \cdot \frac{R}{2} \cdot t = 2 \cdot \frac{V^2}{R} \cdot t = 4 \cdot \frac{V^2}{2R} \cdot t$
Comparing $H_s$ and $H_p$, we see that $H_p = 4 H_s$, meaning the heat produced in the parallel case is 4 times that in the series case. So, the ratio of heat produced in series to parallel is $1:4$.
Therefore, the correct answer is (c) (1: 4).
How is a voltmeter connected in the circuit to measure the potential difference between two points? (80 words)
A voltmeter is designed to measure the potential difference between two points in a circuit. To accurately measure this difference, the voltmeter must be connected in parallel with the component or section of the circuit across which the potential difference is to be measured. This configuration allows the voltmeter to measure the voltage drop directly across the component without significantly altering the current flowing through that part of the circuit. It is crucial to ensure that the voltmeter has a high internal resistance to minimize its impact on the circuit's operation.
A copper wire has diameter $0.5 \mathrm{~mm}$ and resistivity of $1.6 \times 10^{-8} \Omega \mathrm{m}$. What will be the length of this wire to make its resistance $10 \Omega$ ? How much does the resistance change if the diameter is doubled?
To calculate the length of a copper wire needed to produce a resistance of $10 \Omega$, we can use the formula for the resistance $R$ of a wire in terms of its resistivity $\rho$, length $L$, and cross-sectional area $A$:
$$ R = \frac{\rho L}{A} $$
The cross-sectional area $A$ of a wire with diameter $d$ is given by the formula for the area of a circle $A = \pi r^2$, where $r$ is the radius of the wire. The radius is half of the diameter, so $r = \frac{d}{2}$.
Given:
Diameter $d = 0.5 \ mm = 0.5 \times 10^{-3} \ m$
Resistivity $\rho = 1.6 \times 10^{-8} \ \Omega m$
Resistance $R = 10 \ \Omega$
First, we will calculate the area $A$:
Radius $r = \frac{d}{2} = \frac{0.5 \times 10^{-3}}{2} = 0.25 \times 10^{-3} \ m$
$A = \pi r^2$
Then, substituting $A$ and the other known values into the resistance formula, we can solve for $L$.
After calculating $L$, we will address how the resistance changes when the diameter is doubled. This involves recalculating the cross-sectional area with the new diameter and applying the same formula.
The cross-sectional area $A$ of the wire is calculated to be approximately $1.9635 \times 10^{-7} \ m^2$.
Utilizing the given formula for resistance, the length $L$ required to achieve a resistance of $10 \ \Omega$ is:
$$ L = 6.25 \times 10^8 \times A $$
Substituting the value of $A$, we find:
$$ L = 6.25 \times 10^8 \times 1.9635 \times 10^{-7} = 122.72 \ m $$
Hence, the length of the wire needed to obtain a resistance of $10 \ \Omega$ is approximately $122.72 \ m$.
Next, let's calculate how the resistance changes if the diameter of the wire is doubled. Doubling the diameter means the radius also doubles, and since the area $A$ is proportional to the square of the radius $(r^2)$, the area is quadrupled. Given that resistance $R$ is inversely proportional to the area $(R \propto \frac{1}{A})$, quadrupling the area results in a resistance that is one-fourth of the original. Thus, if the original resistance is $10 \ \Omega$, the new resistance with the diameter doubled would be:
$$ R_{new} = \frac{10 \ \Omega}{4} = 2.5 \ \Omega $$
Therefore, if the diameter of the wire is doubled, the resistance of the wire is reduced to $2.5 \ \Omega$.
The values of current $I$ flowing in a given resistor for the corresponding values of potential difference $V$ across the resistor are given below -
$I$ (amperes) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |
$V$ (volts) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |
Plot a graph between $V$ and $I$ and calculate the resistance of that resistor.
The graph of the voltage $(V)$ versus current $(I)$ from the given data points, as well as the linear fit to these points are presented below:
The linear fit equation derived from these data points is:
$$ V = 3.32805I + 0.0310976 $$
This equation is in the form of Ohm’s law, $V = IR$, where $R$ represents the resistance. Comparing the linear fit equation with Ohm’s law, we can identify the slope of the line $3.32805$ as the resistance $(R)$ of the resistor.
Therefore, the resistance of the resistor is 3.32805 ohms.
When a $12 \mathrm{~V}$ battery is connected across an unknown resistor, there is a current of $2.5 \mathrm{~mA}$ in the circuit. Find the value of the resistance of the resistor.
To find the value of the resistance $(R)$ of the resistor when a voltage $(V)$ is applied across it and a current ((I)) flows through it, you can use Ohm's Law, which is given by:
$$ R = \frac{V}{I} $$
Given that (V = 12) V (volts) and (I = 2.5) mA (milliamperes), first we need to convert the current from milliamperes to amperes because the standard unit of current in the International System of Units (SI) is the ampere (A).
$1 mA$ = $1*10^{-3}$ A, therefore:
$$ I = 2.5 , \text{mA} = 2.5*10^{-3} , \text{A} $$
Now, substituting the given values into Ohm's Law:
$$ R = \frac{12 , \text{V}}{2.5*10^{-3} , \text{A}} $$
Let's calculate the resistance $R$. The value of the resistance of the resistor is 4800 ohms $(4800 , \Omega)$.
A battery of $9 \mathrm{~V}$ is connected in series with resistors of $0.2 \Omega, 0.3 \Omega, 0.4 \Omega, 0.5 \Omega$ and $12 \Omega$, respectively. How much current would flow through the $12 \Omega$ resistor?
When resistors are connected in series, the total resistance $(R_{total})$ is the sum of all individual resistances. Let's calculate the total resistance first:
$$ R_{total} = 0.2\ \Omega + 0.3\ \Omega + 0.4\ \Omega + 0.5\ \Omega + 12\ \Omega $$
Adding these together gives:
$$ R_{total} = 13.4\ \Omega $$
Given a battery voltage $(V)$ of $9V$, we can find the current $(I)$ using Ohm's Law, which states $V = IR$. Rearranging this gives $I = \frac{V}{R}$.
Substituting the given values yields:
$$ I = \frac{9V}{13.4\ \Omega} $$
Calculating this gives:
$$ I = \frac{9}{13.4} \approx 0.6716\ \text{A} $$
Therefore, the current that would flow through the $12\ \Omega$ resistor, as well as through any other component in this series circuit, is approximately $0.6716\ \text{A}$ or $671.6\ \text{mA}$.
How many $176 \Omega$ resistors (in parallel) are required to carry $5 \mathrm{~A}$ on a $220 \mathrm{~V}$ line?
To determine how many $176 \Omega$ resistors $in parallel$ are required to carry $5 \mathrm{~A}$ on a $220 \mathrm{~V}$ line, we first need to understand the basic principles involved:
Ohm's Law: $$V = I \times R$$ where $V$ is the voltage across the resistor, $I$ is the current flowing through the resistor, and $R$ is the resistance.
Resistors in Parallel: The total resistance $(R_{\text{total}})$ of resistors connected in parallel is given by: $\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}$ For $n$ resistors of the same value $(R)$, this simplifies to: $\frac{1}{R_{\text{total}}} = \frac{n}{R}$ or equivalently, $R_{\text{total}} = \frac{R}{n}$
Given these principles, we can solve the problem as follows:
First, we need to calculate the total resistance needed to carry $5 \mathrm{~A}$ using the $220 \mathrm{~V}$ line. Using Ohm's Law $(V = I \times R)$, we solve for $R): $$R = \frac{V}{I}$
Substituting the given values: $$R = \frac{220 \mathrm{~V}}{5 \mathrm{~A}} = 44 \Omega$$
This is the total resistance required to carry $5 \mathrm{~A}$ on a $220 \mathrm{~V}$ line.
Next, we use the formula for resistors in parallel to find out how many resistors we need. Since we know $R_{\text{total}}$ should be $44 \Omega$ and each resistor has a resistance of $176 \Omega$, we calculate $n$ (the number of resistors in parallel) using: $$R_{\text{total}} = \frac{R}{n}$$
Rearranging for $n$: $$n = \frac{R}{R_{\text{total}}}$$
Substituting the values: $$n = \frac{176}{44} = 4$$
Therefore, 4 resistors of $176 \Omega$ each in parallel are required to carry $5 \mathrm{~A}$ on a $220 \mathrm{~V}$ line.
Show how you would connect three resistors, each of resistance $6 \Omega$, so that the combination has a resistance of (i) $9 \Omega$, (ii) $4 \Omega$.
To achieve different total resistances by combining three resistors, we can use series and parallel arrangements. Each resistor has a resistance of (6 \Omega).
(i) $9 \Omega$
To get a total resistance of $9 \Omega$consider the following approach:
Connect two resistors in parallel. The equivalent resistance $(R_{\text{parallel}})$ of two resistors, each having resistance $R$, connected in parallel is given by: $$ \frac{1}{R_{\text{parallel}}} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} $$ For two $6 \Omega$ resistors in parallel, $$ \frac{1}{R_{\text{parallel}}} = \frac{2}{6 \Omega} = \frac{1}{3 \Omega} $$ So, $R_{\text{parallel}} = 3 \Omega$.
Next, connect this parallel combination in series with the third resistor. The total resistance in series $(R_{\text{series}})$ is simply the sum of all individual resistances. $$ R_{\text{series}} = R_{\text{parallel}} + R_{3} = 3 \Omega + 6 \Omega = 9 \Omega $$
So, to achieve a total resistance of $9 \Omega$, connect two resistors in parallel, and then this combination in series with the third resistor.
(ii) $4 \Omega$
For a total resistance of $4 \Omega$, use the following configuration:
First, connect two resistors in series. The resistance $R_{\text{series-2}}$ of two resistors, each having resistance $R$, connected in series is given by: $$ R_{\text{series-2}} = 2R = 2 \times 6 \Omega = 12 \Omega $$
Next, connect this series combination in parallel with the third resistor. The equivalent resistance $(R_{\text{total}})$ for this parallel setting can be found using the formula for parallel resistance: $$ \frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{series-2}}} + \frac{1}{R_{3}} $$ Here, $R_{3} = 6 \Omega$ and $R_{\text{series-2}} = 12 \Omega$, $$ \frac{1}{R_{\text{total}}} = \frac{1}{12 \Omega} + \frac{1}{6 \Omega} = \frac{1+2}{12 \Omega} = \frac{1}{4 \Omega} $$ Therefore, $R_{\text{total}} = 4 \Omega$.
Thus, for a $4 \Omega$ total resistance, connect two resistors in series and then connect this series combination in parallel with the third resistor.
Several electric bulbs designed to be used on a $220 \mathrm{~V}$ electric supply line, are rated $10 \mathrm{~W}$. How many lamps can be connected in parallel with each other across the two wires of $220 \mathrm{~V}$ line if the maximum allowable current is $5 \mathrm{~A}$ ?
To solve this problem, we first need to calculate the current drawn by each electric bulb when it is connected across a $220 , \text{V}$ electric supply line.
The power rating of each bulb $(P)$ is $10 , \text{W}$, and the voltage $(V)$ is $220 , \text{V}$. The power rating can be used to calculate the current $(I)$ using the formula $P = VI$, where $I$ is the current in amperes $A$.
Rearranging the formula to solve for $I$we get:
$$ I = \frac{P}{V} $$
Substituting $P = 10 , \text{W}$ and $V = 220 , \text{V}$, we can find $I$:
$$ I = \frac{10}{220} $$
Let's calculate the current drawn by one bulb:
$$ I = \frac{10 , \text{W}}{220 , \text{V}} = \frac{1}{22} , \text{A} $$
Now, the maximum allowable current for the entire circuit is $5 , \text{A}$. To find out how many bulbs can be connected in parallel without exceeding the maximum allowable current, we divide the total allowable current by the current drawn by one bulb:
$$ \text{Number of bulbs} = \frac{5 , \text{A}}{\frac{1}{22} , \text{A}} $$
Let's calculate the number of bulbs:
$$ \text{Number of bulbs} = 5 \times 22 = 110 $$
Therefore, 110 bulbs can be connected in parallel across the two wires of the $220 , \text{V}$ line without exceeding the maximum allowable current of $5 , \text{A}$.
A hot plate of an electric oven connected to a $220 \mathrm{~V}$ line has two resistance coils A and B, each of $24 \Omega$ resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
To find the currents in the three cases (using the coils separately, in series, and in parallel), we'll use Ohm's law and the formulas for total resistance in series and parallel circuits. Ohm's law is given by:
$$ I = \frac{V}{R} $$
where $I$ is the current (in amps, A), $V$ is the voltage $in volts, V$, and $R$ is the resistance $in ohms, (\Omega)$.
1. Using the Coils Separately
When using the coils separately (assuming this means using one coil at a time), each coil has a resistance of $24 \Omega$. The voltage is $220V$.
For each coil:
$$ I = \frac{220V}{24\Omega} $$
Calculating the current for a single coil:
$$ I = \frac{220}{24} \approx 9.17 , A$$
2. Using the Coils in Series
When the coils are used in series, the total resistance, (R_{\text{total}}), is the sum of the individual resistances:
$$ R_{\text{total}} = R_A + R_B = 24\Omega + 24\Omega = 48\Omega $$
The voltage is $220V$.
The current in the series circuit is:
$$ I = \frac{220V}{48\Omega} $$
Calculating the current:
$$ I = \frac{220}{48} \approx 4.58 , A$$
3. Using the Coils in Parallel
For parallel circuits, the total resistance, $R_{\text{total}}$, can be found using the formula for two resistors in parallel:
$$ \frac{1}{R_{\text{total}}} = \frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{24\Omega} + \frac{1}{24\Omega} = \frac{2}{24\Omega} $$
$$ R_{\text{total}} = \frac{24\Omega}{2} = 12\Omega $$
The voltage is $220V$.
The current in the parallel circuit is:
$$ I = \frac{220V}{12\Omega} $$
Calculating the current:
$$ I = \frac{220}{12} \approx 18.33 , A$$
To summarize, the currents in the three cases are:
Using the coils separately: approximately $9.17 , A$ each.
Using the coils in series: approximately $4.58 , A$.
Using the coils in parallel: approximately $18.33 , A$.
Compare the power used in the $2 \Omega$ resistor in each of the following circuits: (i) a $6 \mathrm{~V}$ battery in series with $1 \Omega$ and $2 \Omega$ resistors, and (ii) a $4 \mathrm{~V}$ battery in parallel with $12 \Omega$ and $2 \Omega$ resistors.
To compare the power used in the $2 , \Omega$ resistor in each circuit, we first need to calculate the current through and voltage across the resistor in each case, and then apply the formula for electrical power, $P = IV$, where $P$ is the power, $I$ is the current, and $V$ is the voltage.
Circuit (i): Series Connection
In a series circuit, the current is the same through all components. The total resistance $(R_{\text{total}})$ is the sum of all resistances:
$$ R_{\text{total}} = R_1 + R_2 = 1 , \Omega + 2 , \Omega = 3 , \Omega $$
Using Ohm's law $(V = IR)$, the total current $(I_{\text{total}})$ from a $6 , V$ battery is:
$$ I_{\text{total}} = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{6 , V}{3 , \Omega} = 2 , A $$
All components have the same current in a series circuit, so the current through the $2 , \Omega$ resistor is $2 , A$. The voltage across it $(V_2)$ is:
$$ V_2 = I_2 R_2 = 2 , A \times 2 , \Omega = 4 , V $$
So, the power used by the $2 , \Omega$ resistor, using $P = IV$, is:
$$ P_1 = I_2 V_2 = 2 , A \times 4 , V = 8 , W $$
Circuit (ii): Parallel Connection
In a parallel circuit, the voltage is the same across all branches. The $2 , \Omega$resistor has a $4 , V$battery across it. Using Ohm's law, the current through the $2 , \Omega$resistor $(I_2)$ is:
$$ I_2 = \frac{V_2}{R_2} = \frac{4 , V}{2 , \Omega} = 2 , A $$
Where $V_2$is the voltage across and $R_2$is the resistance of the $2 , \Omega$resistor. As the voltage across it is $4 , V$(due to the parallel connection with the battery), we directly calculate the power used by the $2 , \Omega$resistor as follows:
$$ P_2 = I_2 V_2 = 2 , A \times 4 , V = 8 , W $$
Comparison
In both circuits, the power used in the $2 , \Omega$resistor is the same, $8 , W$.
Two lamps, one rated $100 \mathrm{~W}$ at $220 \mathrm{~V}$, and the other $60 \mathrm{~W}$ at $220 \mathrm{~V}$, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is $220 \mathrm{~V}$ ?
To solve this problem, we need to calculate the total current drawn from the electric mains supply by the two lamps when they are connected in parallel. The formula to calculate power $(P)$in terms of voltage $(V)$and current $(I)$is:
$$ P = V \times I $$
Therefore, to find the current drawn by each lamp, we can rearrange the formula to solve for (I):
$$ I = \frac{P}{V} $$
Let's calculate the current drawn by each lamp:
For the 100 W lamp:
$$ I_1 = \frac{100 , \text{W}}{220 , \text{V}} $$
For the 60 W lamp:
$$ I_2 = \frac{60 , \text{W}}{220 , \text{V}} $$
Finally, since the lamps are connected in parallel, the total current ((I_{\text{total}})) drawn from the mains is the sum of the currents drawn by each lamp:
$$ I_{\text{total}} = I_1 + I_2 $$
Let's calculate these values. The current drawn by each lamp is calculated as follows:
For the 100 W lamp: ( I_1 = 0.4545 , \text{A} ) (454.5 mA)
For the 60 W lamp: ( I_2 = 0.2727 , \text{A} ) (272.7 mA)
The total current $(I_{\text{total}})$ drawn from the mains when both lamps are connected in parallel is the sum of $I_1 $and $I_2 $:
$$ I_{\text{total}} = 0.4545 , \text{A} + 0.2727 , \text{A} = 0.7272 , \text{A} $$
Thus, 0.7272 amperes of current is drawn from the line when both lamps are connected in parallel to an electric mains supply with a voltage of $220 , \text{V}$.
Which uses more energy, a $250 \mathrm{~W} \mathrm{TV}$ set in $1 \mathrm{hr}$, or a $1200 \mathrm{~W}$ toaster in 10 minutes?
To calculate which uses more energy, we need to understand that energy $E$is the product of power $P$ and time $t$, represented as $E = P \times t$.
Energy used by the TV set in 1 hour (60 minutes):
Given:
Power, $P = 250 \mathrm{~W}$
Time, $t = 1 \mathrm{~hour} = 60 \mathrm{~minutes}$
Using the formula $E = P \times t$, $$ E_{\text{TV}} = 250 \mathrm{~W} \times 60 \mathrm{~minutes} = 15,000 \mathrm{~W \cdot minutes} $$
Since energy is often measured in joules $(\mathrm{J})$, and $1 \mathrm{~W} = 1 \mathrm{~J/s}$, we convert the time in minutes to seconds. $$ E_{\text{TV}} = 250 \mathrm{~W} \times 60 \times 60 \mathrm{~seconds} = 900,000 \mathrm{~J} $$
Energy used by the toaster in 10 minutes:
Given:
Power, $P = 1200 \mathrm{~W}$
Time, $t = 10 \mathrm{~minutes}$
Using the formula (E = P \times t), $$ E_{\text{toaster}} = 1200 \mathrm{~W} \times 10 \mathrm{~minutes} = 12,000 \mathrm{~W \cdot minutes} $$
Converting the time in minutes to seconds, $$ E_{\text{toaster}} = 1200 \mathrm{~W} \times 10 \times 60 \mathrm{~seconds} = 720,000 \mathrm{~J} $$
Comparison:
The energy used by the TV in 1 hour is $900,000 \mathrm{~J}$.
The energy used by the toaster in 10 minutes is $720,000 \mathrm{~J}$.
Conclusion: The TV set uses more energy in 1 hour compared to the toaster in 10 minutes.
An electric heater of resistance $8 \Omega$ draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
The rate at which heat is developed in an electric heater can be calculated using the formula for power, which is given by:
$$ P = I^2R $$
where:
$P$ is the power in watts, $W$,
$I$ is the current in amperes, $A$,
$R$ is the resistance in ohms, $\Omega$.
Given:
The resistance $R = 8 \Omega$,
The current $I = 15$ A.
Substituting the given values into the formula gives:
$$ P = (15)^2 \times 8 $$
Let's calculate it. The rate at which heat is developed in the heater is 1800 watts $W$ or 1,800 joules per second $J/s$.
Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Let's address each of your questions:
(a) Why is tungsten used almost exclusively for the filament of electric lamps?
Tungsten is used almost exclusively for the filament of electric lamps due to several of its properties:
High melting point: Tungsten has a melting point of about $3422^{\circ}C$, which is higher than that of most metals. This allows the filament to get extremely hot and emit light without melting.
Good electrical conductivity: While not as good as metals like copper or silver, tungsten’s electrical conductivity is sufficient to make it an effective filament material.
High tensile strength: Tungsten has strong tensile strength, even at high temperatures, which prevents the filament from breaking easily.
Low vapor pressure: At the high temperatures required for lighting, tungsten has relatively low vapor pressure, reducing the rate at which it evaporates. This prolongs the life of the filament by reducing thinning and breakage.
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
Alloys are used in electric heating devices for several reasons:
Higher resistance: Alloys typically have higher electrical resistance than pure metals, which makes them more efficient at converting electrical energy into heat.
Durability: Alloys can be tailored to have higher melting points and greater strength at high temperatures, making them more durable and less likely to break down or deform with repeated heating and cooling cycles.
Cost-effectiveness: Some alloys that have the desired electrical and thermal properties can be less expensive or more readily available than pure metals with similar properties.
(c) Why is the series arrangement not used for domestic circuits?
Series arrangement is not used for domestic circuits because:
Unequal Voltage Drop: In a series circuit, voltage is divided among the appliances based on their resistance. This can result in insufficient voltage for some appliances, causing them not to work properly.
Single Failure Affects All: If one appliance or component in a series circuit fails or is turned off, the circuit is broken, and all devices in the circuit will stop working.
Difficulty in Controlling: It’s difficult to control each appliance individually without affecting others in a series circuit.
(d) How does the resistance of a wire vary with its area of cross-section?
The resistance ((R)) of a wire is inversely proportional to its cross-sectional area ((A)), as determined by the equation:
$$ R = \frac{\rho L}{A} $$
Where:
$R$ is the resistance,
$\rho$ is the resistivity of the material,
$L$ is the length of the wire,
$A$ is the cross-sectional area.
This means that as the area of cross-section increases, the resistance decreases. A thicker wire has less resistance than a thinner wire of the same material and length.
(e) Why are copper and aluminium wires usually employed for electricity transmission?
Copper and aluminium wires are commonly used for electricity transmission due to their:
High Electrical Conductivity: Both metals have excellent electrical conductivity, meaning they can carry more current than most other materials.
Durability: They both withstand environmental exposure quite well, which is important for transmission lines.
Flexibility: Both metals are ductile, meaning they can be drawn into thin wires and are flexible enough to be routed as needed.
Cost: While copper is more expensive, it is highly efficient. Aluminium offers a good balance of conductivity and cost, especially for high-voltage or long-distance transmission lines where the less weight of aluminium is an advantage.
Each material offers a balance of properties ideally suited for the purposes of efficient and reliable electrical transmission.
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Ask Chatterbot AIExtra Questions and Answers - Electricity | NCERT | Science | Class 10
When a $12 \mathrm{~V}$ battery is connected across an unknown resistor, there is a current of 2.5 $\mathrm{mA}$ in the circuit. Find the value of the resistance of the resistor.
A) $4.8 \Omega$
B) $50 \Omega$
C) $4800 \Omega$
D) $5000 \Omega$
Using Ohm's Law, which is represented by the formula: $$ V = IR $$ Here, the voltage ($V$) is given as $12 \text{ V}$, and the current ($I$) is $2.5 \text{ mA}$.
First, converting the current from milliamperes to amperes: $$ I = 2.5 \text{ mA} = 2.5 \times 10^{-3} \text{ A} $$
Now, we can solve for the resistance ($R$) using the formula for Ohm's Law: $$ R = \frac{V}{I} = \frac{12 \text{ V}}{2.5 \times 10^{-3} \text{ A}} = 4800 \Omega $$
Thus, the value of the resistance is $4800 \Omega$, corresponding to option C.
The voltage across each appliance connected in series is different.
A) True
B) False
The correct option is A) True
Voltage division occurs in a series circuit, leading to different voltages across each appliance.
A piece of wire of resistance $R$ is cut into five equal parts, keeping the same cross-sectional area. What will be the new resistance of each resistor?
(A) $\frac{R}{5}$
(B) $5R$
(C) $R$
(D) $\frac{R}{2}$
The correct answer is (A) $\frac{R}{5}$.
Resistance ($R$) is directly proportional to the length of the wire considering the uniform cross-sectional area. When the wire is cut into five equal parts, each part will have a length equal to $\frac{1}{5}$th of the original length. Hence, the resistance of each part will be:
$$ \frac{1}{5} \times R = \frac{R}{5} $$
Thus, the resistance of each of the five equal parts of the wire is $\frac{R}{5}$.
If the resistance of a conductor becomes nearly equal to zero, then such conductors are called_ $\qquad$
A non-conductors
B resistive conductors
C ohmic conductors
D superconductors
The correct answer is D) superconductors.
When the resistance of a conductor approaches zero, it is called a superconductor. This type of material has the characteristic that its electrical resistance vanishes and allows current to flow without any energy loss, effectively giving it infinite conductivity.
If a wire of resistance $20 \Omega$ is covered with ice and a voltage of $210 \mathrm{~V}$ is applied across the wire, what will be the rate of melting of ice? [Specific latent heat of fusion of ice = $80 \mathrm{cal/g}$]
A) $5.85 \mathrm{~g/s}$ B) $1.92 \mathrm{~g/s}$ C) $6.60 \mathrm{~g/s}$ D) $5.65 \mathrm{~g/s}$
The correct option is C
$$
6.60 , \text{g/s}
$$
First, we denote:
Mass of ice melting per second as $m$ grams.
Resistance of the wire as $R = 20 , \Omega$.
Voltage applied as $V = 210 , \mathrm{V}$.
Specific latent heat of fusion of ice, $L = 334 , \mathrm{J/g}$ (equivalent to the given $80 , \mathrm{cal/g}$ if calculated with correct conversion factor).
The power dissipated in the wire, $P$, is calculated using: $$ P = \frac{V^2}{R} $$ Substituting the values, we get: $$ P = \frac{(210)^2}{20} = 2205 , \mathrm{W} $$
The heat produced by the wire, $H$, over $t$ seconds is: $$ H = P \cdot t $$
For melting ice, the amount of heat required, $Q$, is given by: $$ Q = m \cdot L $$
Setting the heat generated equal to the heat required for melting, and considering a time interval of $t = 1 , s$: $$ \frac{V^2}{R} \times 1 = m \cdot 334 $$
Solving for $m$: $$ m = \frac{2205}{334} \approx 6.60 \text{ g/s} $$
Thus, the rate of ice melting is approximately 6.60 grams per second.
The most significant source of energy in the solar system is:
A) Sun
B) Moon
C) Stars
D) Petroleum
The correct option is A) Sun.
The Sun is the most significant source of energy in our solar system. It supports all life on our planet through various means, primarily by providing light and warmth.
Electricity is a primary form of energy. State true or false.
A) True
B) False
The correct answer is B) False.
Electricity is a secondary form of energy because it must be produced using other primary sources of energy.
Electricity can be used to replace the fuel and power a rocket (using today's mechanism) just like an electric car?
A. True, as electricity is also a form of energy.
B. False, as electricity is not a source of energy.
C. False, as the rocket needs to throw out fuel in the opposite direction to move.
D. True, as the mechanism for a car and a rocket is similar.
The correct option is C. False, as the rocket needs to throw out fuel in the opposite direction to move.
Rocket propulsion is based on Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. In the context of a rocket, it ejects its fuel at high speed in one direction, generating the necessary thrust (reaction) that propels the rocket in the opposite direction. Electric cars, on the other hand, rely on electric motors that convert electrical energy into mechanical energy, which is not applicable to creating the massive thrust needed for space travel through the same mechanism.
Along with live wire and neutral wire, a third wire is also used in domestic electric wiring. What name is given to this third wire?
The name given to the third wire in domestic electric wiring is the Earth Wire. This designation is because the earth wire is typically connected to a metal plate that is buried deep in the earth near the house. It serves as a crucial safety measure and does not influence the power supply in any way.
It is dangerous to connect the switch in the neutral wire. Explain your answer.
Connecting a switch in the neutral wire is considered hazardous due to the following reasons:
Disconnection of Danger: Generally, switches are installed on the live wire. When a switch is opened, it should effectively disconnect or isolate the appliance from the high voltage live wire. This ensures that all parts of the appliance are safe to touch and there is no risk of an electric shock.
Continuous Live Connection: Placing the switch in the neutral wire does not break the circuit from the high voltage source. The appliance remains connected to the live wire even when the switch is off. This situation can lead to potential accidents.
Risk of Electric Shock: If the switch is in the neutral wire, any person touching the appliance or any of its internal components might still come into contact with the live wire. This can result in a dangerous electric shock if the person inadvertently touches the open, live wire believing the circuit is safe and inactive.
Hence, always ensuring the switch is on the live wire is crucial for safety to effectively isolate the electrical device and prevent the risk of electric shocks.
What is electric energy?
Electric energy, also known as electrical energy, is the form of energy that is created by the movement of electrons through a conductor, and it can be used to perform work, such as providing power to electrical appliances. When electrons flow through a conductor, such as a copper wire, they create an electric current. This flow of electric current is what transmits energy from one point to another, enabling the functioning of electrical devices.
For more detailed information, it's beneficial to look up recognized educational resources or specific energy-focused materials.
What should be the rating of an electric fuse for a microwave oven, which is operating on a current rating of 5A?
Option 1: Zero
Option 2: 5A
Option 3: 6A
Option 4: Anything above 5A
The correct option is B: 5A
The rating of an electric fuse should be 5A. In the event that the circuit gets overloaded beyond 5A, which matches the oven's current rating, the electric fuse will burn out and disconnect the circuit, thereby protecting the appliance and preventing potential damage or hazards.
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