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The arrangement of resistors shown in the below figure is connected to a battery.

The arrangement of resistors shown in the below figure is connected to a battery. CBSE 2026 Science Class 10 Sample Question Paper Solved.

The power dissipation in the 100 Ω resistor is 81 W. Calculate

(a) the current in the circuit

(b) the reading in the voltmeter V2

(c) the reading in the voltmeter V1

Current Electricity

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Answer

Given,

Power dissipated in 100 Ω resistor (P\text P) = 81 W

(a) Let, current in the circuit be 'I\text I'.

Then,

P=I2R⇒I2=P100=81100⇒I2=0.81⇒I=0.81⇒I=0.9 A\text P = \text I^2 \text R \\[1em] \Rightarrow \text I^2 = \dfrac{\text P}{100} \\[1em] = \dfrac{81}{100} \\[1em] \Rightarrow \text I^2 = 0.81 \\[1em] \Rightarrow \text I = \sqrt {0.81} \\[1em] \Rightarrow \text I = 0.9\ \text A

Hence, the current in the circuit is 0.9 A.

(b) From the figure, two 25 Ω resistors are in parallel and their effective resistance is given by,

1RP=125+125⇒1RP=225⇒RP=252⇒RP=12.5 Ω\dfrac{1}{\text R_ \text P} = \dfrac{1}{25} + \dfrac{1}{25} \\[1em] \Rightarrow \dfrac{1}{\text R_ \text P} = \dfrac{2}{25} \\[1em] \Rightarrow \text R\text P = \dfrac{25}{2} \\[1em] \Rightarrow \text R\text P = 12.5\ \text Ω

The reading in the voltmeter V2 is same as the potential difference across RP\text R_\text P which is given by,

V2=IRP=0.9×12.5⇒V2=11.25 V\text V2 = \text {IR}\text P \\[1em] = 0.9\times 12.5 \\[1em] \Rightarrow \text V_2 = 11.25\ \text V

Hence, the reading in the voltmeter V2 is 11.25 V.

(c) Now, RP\text R_ \text P is in series with 100 Ω resistor and their effective resistance is given by,

RS=100+12.5⇒RS=112.5 Ω\text R_ \text S = 100 + 12.5 \\[1em] \Rightarrow \text R_\text S = 112.5\ \text Ω

The reading in the voltmeter V1 is same as the voltage supply of the battery which is given by,

V1=IRS=0.9×112.5⇒V1=101.25 V\text V1 = \text {IR}\text S \\[1em] = 0.9\times 112.5 \\[1em] \Rightarrow \text V_1 = 101.25\ \text V

Hence, the reading in the voltmeter V1 is 101.25 V.

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