Current Electricity

JEE Main level, numerical answer, medium

339 of 380

Question 339 of 380

As shown in the figure the voltmeter reads 2V2V across 5Ω5\Omega resistor. The resistance of the voltmeter is _____ Ω\Omega

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Answer: 20

Let the resistance of the voltmeter be RΩR\Omega. The voltmeter is in parallel with the 5Ω5\Omega. So the total resistance in the circuit is
(5R5+R+2)Ω\left(\dfrac{5R}{5+R}+2\right)\Omega
Applying Ohm's law in the whole circuit, V=IRV=IR,
I=(32+5R5+R)AI=\left(\dfrac{3}{2+\dfrac{5R}{5+R}}\right)A
It is given that the potential difference across 5Ω5\Omega is 2V2V.
Thus, by Ohm's law,
(32+5R5+R)×(5RR+5)=2⇒15+3R10+7R×5R5+R=2⇒15R=20+14R⇒R=20Ω\left(\dfrac{3}{2+\dfrac{5R}{5+R}}\right)\times \left(\dfrac{5R}{R+5}\right)=2 \Rightarrow \dfrac{15+3R}{10+7R}\times \dfrac{5R}{5+R}=2 \Rightarrow 15R=20+14R \Rightarrow R=20\Omega

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