A battery of emf 12 V and internal resistance 4 ohm is connected to a resistor. If the current in the circuit is 0.6 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Emf of the battery, E = 12V Internal resistance of the battery, r = 4 ohm Current = 0.6 A so, I = E/(R + r) R + r = E/I Â = 12/0.6 = 20 ohm Therefore = 20 - 4 = 16 ohm Terminal voltage of resistor = V so, V = IR => 0.6 x 16 => 9.6 V
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