JAMB - Physics (1995 - No. 32)
A 12V battery has an internal resistance of 0.5W.If a cable of 1.0W resistance is connected across the two terminals of the battery, the current drawn from the battery is
16.0A
8.0A
0.8A
0.4A
Explanation
I = \(\frac{E.m.f}{R + r}\)
= \(\frac{12}{1 + 0.5}\)
I = 8A
= \(\frac{12}{1 + 0.5}\)
I = 8A
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