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.
- 13.0V
-
11.0V
- 3.0V
- 1.0V
The correct answer is: B
Explanation
\(\frac{V}{R} = \frac{E}{R + r} = \frac{V}{5.5} = \frac{12}{6}\)v = 11.0V