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A battery of the emf 18 V and internal resistance of 1omega are connected as shown in figure. Then, the voltmeter reading is |
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Answer» 10 V = 18 - 10 - 8 Net resistance of the CIRCUIT = 3 + 1 = 4 Current in the circuit I `= (8)/(4) = 2A` `therefore` Potential difference (V) = E - ir = 18 - 6 12 V |
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