In a supply line of 100 V there is a resistance of `1000 Omega`. In between our terminal of the resistance and its mid point, a voltmeter is connected which gives a reading of 40 V. determine the resistance of the voltmeter.

Explanation

Suppose,resistance of the voltmeter is R, In fig.1.23, C is the midpoint of the resistance AB.
So, resistance of each portion AC and BC=`1000/2=500 Omega`
Since the reading of the voltmeter =40 V,
`thereforeV_A-V_C=40V,soV_C-V_B=100-40=60V`
Now, main current of the circuit,
`I=(V_C-V_B)/(resistance of BC)=60/500=0.12A`
Again, current in the resistance AC,
`I'=(V_A-V_C)/(resistance of AC)`
`=40/500=0.08A`
So, current in the voltmeter,
`I_V=I-I'=0.12-0.08=0.04A`
Therefore,resistance of the voltmeter,
`R=(V_A-V_C)/I_V=40/0.04=1000Omega`