Question on: WAEC Physics - 2006
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
A
0.25A
B
0.30A
C
0.35A
D
0.50A
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Correct Option: A
I = (\frac{E}{R + r} = \frac{1.5}{5 + 1})
= (\frac{1.5}{6} = 0.25A)
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