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.
-
0.63A
- 0.50A
- 0.31A
- 0.22A
The correct answer is: A
Explanation
I = \(\frac{f}{n} = \frac{1}{6 - 2}\)
I = \(\frac{v}{R + r} = \frac{2}{3} + \frac{1}{6} = 0.63\)A