Question 6A323

Transformers operate on the principle of mutual inductance, which requires a *changing* magnetic field to induce a voltage in the secondary winding. This changing field is produced by alternating current (AC). When operating with AC, the primary winding of a transformer presents a high *impedance*, primarily due to its inductive reactance ($X_L = 2\pi f L$). This impedance significantly limits the current drawn from the supply. However, when connected to direct current (DC), the frequency ($f$) is zero. Consequently, the inductive reactance ($X_L$) becomes zero. The primary winding then only presents its very low DC resistance. Applying 110V DC across this low resistance results in an extremely high current flow (Ohm's Law: $I = V/R_{DC}$). This excessive current causes rapid overheating of the primary winding ($P = I^2R$), leading it to burn out. This severe primary overload and potential shorting can also cause damage or burnout to the secondary winding. While a circuit breaker *should* open to protect the circuit, the transformer itself will experience destructive current, leading to damage to either its primary or secondary windings.