A non-ideal step-down transformer connected to an AC mains supply of 200 V operates a 10 V, 20 W lam

Question

A non-ideal step-down transformer connected to an AC mains supply of 200 V operates a 10 V, 20 W lamp at its secondary. If the efficiency of the transformer is 80%, what is the current drawn in the primary circuit?

Professor PC Thomas Classes · Answer

(b) : The output power $P_{out} = 20$ W. Since the transformer is 80% efficient ($\eta = 0.8$), the input power $P_{in}$ is calculated by $P_{in} = P_{out} / \eta = 20 / 0.8 = 25$ W. We also know that $P_{in} = V_p I_p$. Given the primary voltage $V_p = 200$ V, the primary current is $I_p = P_{in} / V_p = 25 / 200 = 0.125$ A.