Consider the diagram shown below. A voltmeter | vedclass

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(C) When a voltmeter of resistance $150\,\Omega$ is connected in parallel across the $100\,\Omega$ resistor between $A$ and $B$,the equivalent resista

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$31$

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(C) When a voltmeter of resistance $150\,\Omega$ is connected in parallel across the $100\,\Omega$ resistor between $A$ and $B$,the equivalent resistance $R_{AB}$ is given by:$R_{AB} = \frac{150 	imes 100}{150 + 100} = \frac{15000}{250} = 60\,\Omega$The total resistance of the circuit $R_{eq}$ is the sum of $R_{AB}$ and the resistance between $B$ and $C$ $(R_{BC} = 100\,\Omega)$:$R_{eq} = R_{AB} + R_{BC} = 60 + 100 = 160\,\Omega$The total current $I$ flowing through the circuit is:$I = \frac{V}{R_{eq}} = \frac{50}{160} = 0.3125\,A$The potential drop across $B$ and $C$ is:$V_{BC} = I 	imes R_{BC} = 0.3125 	imes 100 = 31.25\,V$Rounding to the nearest integer,the potential drop is $31\,V$.

Consider the diagram shown below. $A$ voltmeter of resistance $150\,\Omega$ is connected across $A$ and $B$. The potential drop across $B$ and $C$ measured by the voltmeter is $...........\,V$.

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