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

A.

0.75\(\Omega\)

B.

1.5\(\Omega\)

C.

0.5\(\Omega\)

D.

1.0\(\Omega\)

E.

0.6\(\Omega\)

View answer & explanation

Correct answer is A

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\(\Omega\)

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