Six dry cells each of e.m.f. 2.0 and internal resistance of 1.0\(\Omega\) are connected in parallel across a load of 3.0\(\Omega\). Calculate the effective current in the circuit.
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Correct Answer: Option A
Explanation:
I = \(\frac{f}{n} = \frac{1}{6 - 2}\)
I = \(\frac{v}{R + r} = \frac{2}{3} + \frac{1}{6} = 0.63\)A
I = \(\frac{f}{n} = \frac{1}{6 - 2}\)
I = \(\frac{v}{R + r} = \frac{2}{3} + \frac{1}{6} = 0.63\)A