A battery of e.m.f. 12.0V and internal resistance0.5\(\Omega\) is connected to 1.5\(\Omega\) and 4.0\(\Omega\) series resistor. Calculate the terminal voltage of the battery.
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Correct Answer: Option B
Explanation:
\(\frac{V}{R} = \frac{E}{R + r} = \frac{V}{5.5} = \frac{12}{6}\)
v = 11.0V
\(\frac{V}{R} = \frac{E}{R + r} = \frac{V}{5.5} = \frac{12}{6}\)
v = 11.0V