# Electricity - Class 10 - Science

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## Back Questions - Electricity | NCERT | Science | Class 10

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) $V^{2} / R$

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Sign up nowAn 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) $25 \mathrm{~W}$

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Sign up nowTwo 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

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Sign up nowHow is a voltmeter connected in the circuit to measure the potential difference between two points? (80 words)

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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?

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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.

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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.

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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?

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How many $176 \Omega$ resistors (in parallel) are required to carry $5 \mathrm{~A}$ on a $220 \mathrm{~V}$ line?

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Show how you would connect three resistors, each of resistance $6 \Omega$, so that the combination has a resistance of (i) $9 \Omega$, (ii) $4 \Omega$.

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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 $5 \mathrm{~A}$ ?

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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?

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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.

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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 $220 \mathrm{~V}$ ?

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Which uses more energy, a $250 \mathrm{~W} \mathrm{TV}$ set in $1 \mathrm{hr}$, or a $1200 \mathrm{~W}$ toaster in 10 minutes?

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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.

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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?

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## Extra Questions - 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

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"It is dangerous to connect the switch in the neutral wire. Explain your answer."