1.

A battery of the emf 18 V and internal resistance of 1omega are connected as shown in figure. Then, the voltmeter reading is

Answer»

10 V
12 V
16 V
8 V

Solution :Net emf of the COMBINATION
= 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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