Three electric cells each of e.m.f 1.5V and internal resistance of 1.0 $\Omega$ are connected in parallel across an external resistance of $\frac{2}{3}$ $\Omega$ . Calculate the value of the current in the resistor

0.5 A

0.9 A

1.5 A

4.5 A

Correct answer is C

E.m.f = 1.5

r = $\frac{1 \times\ 1 \times 1}{(1 \times\ 1) + (1 \times\ 1) + (1 \times\ 1)}$

r = $\frac{1}{3}$ $\Omega$

I = $\frac{E.m.f}{r + R}$

I = $\frac{1.5}{\frac{1}{3}\ + \frac{2}{3}}$

I = 1.5A

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