In a step-down transformer, the primary coil has 2000 turns and the secondary coil has 400 turns. If the primary voltage is 240 V, what is the current in the secondary coil when the primary current is 0.5 A? (Assume 100% efficiency.)

For an ideal transformer, $\frac{N_p}{N_s} = \frac{V_p}{V_s} = \frac{I_s}{I_p}$. The turns ratio is $\frac{2000}{400} = 5$. Therefore $I_s = I_p \times \frac{N_p}{N_s} = 0.5 \times 5 = 2.5\,\text{A}$. Alternatively, using power conservation: $P = V_p I_p = 240 \times 0.5 = 120\,\text{W}$; $V_s = \frac{240}{5} = 48\,\text{V}$; $I_s = \frac{120}{48} = 2.5\,\text{A}$.