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In the circuit shown below, each battery is 5V and has an internal resistance of 0.2 ohm. The reading in the ideal voltmeter V is . |
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Answer» Let a current I flow through the circuit. Net emf of the circuit `= 8(5V) = 40V` Net resistance in the circuit `= 8(0.2 Omega) = 1.6Omega` Current flowing through the circuit, `I = (40 V)/(1.6Omega) = 25 A` The voltmeter reading would be `V = E-IR = (5V)-(25A)(0.2Omega)` `=5V-5V = 0`. |
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