The power cube transformer for a portable CD player has an output of 4.6 V and 600 mA when plugged into a 120V per 60Hz outlet What is the peak current in the primary coil?

To find the peak current in the primary coil, we can use the power equation. The output power is (P = V \times I = 4.6 , \text{V} \times 0.6 , \text{A} = 2.76 , \text{W}). Assuming the transformer is ideal and neglecting losses, the input power in the primary coil is also 2.76 W. The primary voltage is 120 V, so the primary current (RMS) is (I_p = \frac{P}{V} = \frac{2.76 , \text{W}}{120 , \text{V}} = 0.023 , \text{A} ) (or 23 mA). The peak current is given by (I_{peak} = I_{rms} \times \sqrt{2} \approx 0.023 , \text{A} \times 1.414 \approx 0.0325 , \text{A} ) (or 32.5 mA).