Question

P, Q is a uniform wire of resistance $2000\Omega$ and M the mid point of PQ. A voltmeter of resistance 1000 $\Omega$ is connected between P and M. The reading of the voltmeter, the potential differences applied between PQ is 150 volt will be :-

• A. 150 volt
• B. 100 volt
• C. 75 volt
• D. 50 volt

Answer 1

The correct option is D 50 volt

Resistance of PM and MQ part is equal to $1000\Omega$ (As M is mid point, $R\propto$length)
The equivalent resistance $R_{eq}=\left(1000||1000\right)+1000=500+1000=1500\Omega$
The current $I=\frac{150}{1500}=0.1A$
The current through PM part is $I/2=0.1/2=0.05$
Thus, voltmeter reading $=$ potential across PM $=0.05\times 1000=50V$