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Q.

In the current carrying branch shown in the figure, the voltmeter has a resistance of 100Ω.If power dissipated in 2Ω,resistance is 18 watt, the reading of the voltmeter is

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a

100 Volt

b

300 Volt

c

400 Volt

d

200 Volt

answer is D.

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Detailed Solution

i2×2=18 ⇒i=3ACurrent through the voltmeter, iv=3×200100+200=2A∴voltmeter reading =2×100 volt =200 volt.
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