JAMB Physics 2020 Paper

Concept:The total resistance of a series-parallel circuit is obtained by simplifying the parallel branch first, then adding the series resistor.Explanation:Total current is $I = 2\ \mathrm{A}$ and supply voltage is $V = 12\ \mathrm{V}$.So total resistance is$$R_T = \frac{V}{I} = \frac{12\ \mathrm{V}}{2\ \mathrm{A}} = 6\ \Omega.$$From the circuit, the $3\ \Omega$ resistor is in series with the parallel combination of $R$ and $6\ \Omega$.Therefore,$$R_T = 3 + \frac{R \times 6}{R + 6}.$$Substitute $R_T = 6\ \Omega$:$$6 = 3 + \frac{6R}{R + 6}.$$Subtract $3$ from both sides:$$3 = \frac{6R}{R + 6}.$$Cross-multiply:$$3(R + 6) = 6R$$$$3R + 18 = 6R$$$$18 = 3R$$$$R = 6\ \Omega.$$Answer:The value of $R$ is $6\ \Omega$.Correct option: B.