Two cells each of e.m.f. 1.5V and an internal resistance 2\(\Omega\) are connected in parallel. Calculate the current flowing when the cells are connected to a 1\(\Omega\) resistor
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Correct Answer: Option A
Explanation:
E = 1.5v
rT = \(\frac{2 \times 2}{2 + 2}\)
= 1
e.m.f. = I(R + r)
1.5 = I(1 + 1)
I = 0.75\(\Omega\)
E = 1.5v
rT = \(\frac{2 \times 2}{2 + 2}\)
= 1
e.m.f. = I(R + r)
1.5 = I(1 + 1)
I = 0.75\(\Omega\)