a cell of e.m.f. 1.5V and internal resistance 1.0\(\Omega\) is connected to two resistor of resistance 2.0\(\Omega\) and 3.0\(\Omega\) in series. Calculate the current through the resistors
-
0.25A
- 0.30A
- 0.35A
- 0.50A
The correct answer is: A
Explanation
I = \(\frac{E}{R + r} = \frac{1.5}{5 + 1}\)= \(\frac{1.5}{6} = 0.25A\)