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
The correct answer is: B
Explanation
I = \(\frac{E.m.f}{R + r}\)= \(\frac{12}{1 + 0.5}\)
I = 8A