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When an ammter of negligible internal resistance is inserted in series with circuit it reads `1 A` When the voltmeter of very large resistance is connected across `X` it reads `1 V`. When the point `A` and `B` are shorted by a conducting wire, the voltmeter meausres `10 V` across the battery. The internal resistance of the battery is equal to A. zeroB. `0.5 Omega`C. `0.2 Omega`D. `0.1 Omega` |
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Answer» Correct Answer - C c. ` I = 1A` r is the internal resistance of the battery. `12= (X+Y+r)(1)` Also `1 = (X)(1) or X = 1Omega` Voltage across X (when A and B are shorted) `10 = (12/(X+r)) X or 10 = (12)/(1+r) or 1 +r = 6/5` or `r = 1/5 Omega` . |
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