1. Suppose that when a 12.0-V battery drives a current of 2.4 A through a resist

Question

1. Suppose that when a 12.0-V battery drives a current of 2.4 A through a resistive device, its terminal voltage is 10.6 V. How much is the internal resistance (in ohms) of the 12.0-V battery?

2. Suppose three identical batteries are connected in series, with the same polarity, in a flashlight, to a bulb with a resistance of 3.6 ohms. The batteries each have an emf of 1.8 volts. If the current flow is 0.62 A, what is the internal resistance (in ohms) of each battery? Neglect the resistance of wires and connectors.

Explanation / Answer

1. If the internal resister is r and the voltage is V that drives a current i through the resister then the terminal voltage will be given Vterminal = V - i*r.

so 10.6 = 12 - 2.4*r => r = 0.5833 ohm.

2. Suppose we want to apply KVL here. So emf1 - i*r +emf2 - i*r +emf3 - i*r -i*R = 0

or 3*emf -3*i*r -i*R = 0;

5.4 = 0.62*(3r+3.6)

r = 1.7 ohms

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