WAEC Physics 2012 Paper

Concept:When identical cells are connected in parallel, the effective e.m.f. is the same as one cell, but the internal resistance is divided by the number of cells.Explanation:Each cell has e.m.f. $E = 2.0\ \mathrm{V}$ and internal resistance $r = 1.0\ \Omega$.Number of cells, $n = 6$.Effective e.m.f. for parallel connection: $E_{eff} = E = 2.0\ \mathrm{V}$.Effective internal resistance: $r_{eff} = \frac{r}{n} = \frac{1.0}{6}\ \Omega$.Load resistance: $R = 3.0\ \Omega$.Total resistance in the circuit: $R_{total} = R + r_{eff}$.Thus, $R_{total} = 3.0 + \frac{1}{6} = \frac{19}{6}\ \Omega$.Using Ohm's law, $I = \frac{E_{eff}}{R_{total}}$.Therefore, $I = \frac{2.0}{\frac{19}{6}} = \frac{12}{19} = 0.63\ \mathrm{A}$.Answer:The effective current in the circuit is $0.63\ \mathrm{A}$. The correct option is A.