Three identical cells each of emf 2V and internal resistance 0.2 ohm are connected in series to an external resistor of 7.4 ohm. Calculate the current in the circuit.

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Solution

Three internal resistance and one external resistance all are connected in series so net resistance of the circuit is
R = 0.2+0.2+0.2+7.4 = 8 ohm

current in the circuit = V/R = 2/8 = 0.25 A

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Three identical cells each of emf 2V and internal resistance 0.2 ohm are connected in series to an external resistor of 7.4 ohm. Calculate the current in the circuit.

Three internal resistance and one external resistance all are connected in series so net resistance of the circuit is R = 0.2+0.2+0.2+7.4 = 8 ohm current in the circuit = V/R = 2/8 = 0.25 A

Three internal resistance and one external resistance all are connected in series so net resistance of the circuit is
R = 0.2+0.2+0.2+7.4 = 8 ohm

current in the circuit = V/R = 2/8 = 0.25 A