WAEC - Physics (2006 - No. 40)
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
Paskaidrojums
I = \(\frac{E}{R + r} = \frac{1.5}{5 + 1}\)
= \(\frac{1.5}{6} = 0.25A\)
= \(\frac{1.5}{6} = 0.25A\)