a cell of e m f 1 5V and internal resistance 1 ... - WAEC Physics 2006 Question
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
correct option: a
I = \(\frac{E}{R + r} = \frac{1.5}{5 + 1}\)
= \(\frac{1.5}{6} = 0.25A\)
= \(\frac{1.5}{6} = 0.25A\)
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