Three cells each of e.m.f. 1.5v and an internal resistance of 1.0\(\Omega\) are connected in parallel across a load resistance of 2.67\(\Omega\). Calculate the current in the load?

A.

0.26A

B.

0.41A

C.

0.50A

D.

0.79A

View answer & explanation

Correct answer is C

E = 1.5v

rT = \(\frac{1 \times 1\times 1}{(1\times1)+ (1\times1) +(1\times1)}\)

E.m. f = I(R + r) = 0.33\(\Omega\)

1.5 = I(2.67 + 0.33)

I = 0.5A

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