(i) Given a potential difference of \(V = 6\ V\), resistors of \(1 \ Ω\) and \(2 \ Ω\) are connected in series. The equivalent resistance of the circuit is \(R = 1 + 2 = 3\ Ω\). According to Ohm’s law, \(V = IR\). With \(V=6\ V\) and \(R=3\ Ω\), the current through the circuit is \(I = \frac{V}{R} = \frac{6}{3} = 2\ A\). Since current is the same in a series circuit, the current flowing through the \(2 \ Ω\) resistor is \(2\ A\). The power consumed by the \(2 \ Ω\) resistor is given by \(P = I^2R\). Substituting the values, \(P = 2^2 \times 2 = 8\ W\).
(ii) For a potential difference of \(V = 4\ V\), resistors of \(12 \ Ω\) and \(2\ Ω\) are connected in parallel. In a parallel circuit, the voltage across each component is the same. Therefore, the voltage across the \(2 \ Ω\) resistor is \(4 \ V\). The power consumed by the \(2\ Ω\) resistor is calculated using \(P=\frac {V^2}{R}\). With \(V = 4\ V\) and \(R = 2\ Ω\), the power is \(P = \frac {4^2}{2} = 8\ W\). Thus, the power used by the \(2\ Ω\) resistor is \(8 \ W\).