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Question

When an ammeter of negligible internal resistance is inserted in series with circuit it reads 1A. When the voltmeter of very large resistance is connected across X it reads 1V. When the point A and B are shorted by a conducting wire, the voltmeter measures 10V across the battery. The internal resistance of the battery is equal to
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A
Zero
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B
0.5 Ω
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C
0.2 Ω
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D
0.1 Ω
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Solution

The correct option is C 0.2 Ω
Initially ammeter measures current 1A through the circuit, when voltmeter measures voltage 1V the resistance is,
R=VI=1Ω
When the point A and B are shorted by a conducting wire, the voltmeter measures 10V across the battery.
Hence the current is
I=101=10A
Now, the voltage across the terminals of the battery is
V=EIr
10=1210r
10r=2
r=0.2Ω

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