$A$ $60\, W$ load is connected to the secondary of a transformer whose primary draws line voltage. If a current of $0.54\, A$ flows in the load,what is the current in the primary coil? Comment on the type of transformer being used.
(0.27 A) Given: Load power $P_{S} = 60\, W$,Secondary current $I_{S} = 0.54\, A$,Primary voltage $V_{P} = 220\, V$ (standard line voltage).
Step $1$: Calculate the secondary voltage $V_{S}$.
$P_{S} = V_{S} I_{S} \implies V_{S} = \frac{P_{S}}{I_{S}} = \frac{60}{0.54} \approx 111.1\, V \approx 110\, V$.
Step $2$: Determine the transformer type.
Since $V_{S} < V_{P}$ $(110\, V < 220\, V)$,the transformer is a step-down transformer.
Step $3$: Calculate the primary current $I_{P}$ using the ideal transformer relation $V_{P} I_{P} = V_{S} I_{S}$.
$I_{P} = \frac{V_{S} I_{S}}{V_{P}} = \frac{60}{220} \approx 0.27\, A$.
Thus,the current in the primary coil is $0.27\, A$ and it is a step-down transformer.
Step $1$: Calculate the secondary voltage $V_{S}$.
$P_{S} = V_{S} I_{S} \implies V_{S} = \frac{P_{S}}{I_{S}} = \frac{60}{0.54} \approx 111.1\, V \approx 110\, V$.
Step $2$: Determine the transformer type.
Since $V_{S} < V_{P}$ $(110\, V < 220\, V)$,the transformer is a step-down transformer.
Step $3$: Calculate the primary current $I_{P}$ using the ideal transformer relation $V_{P} I_{P} = V_{S} I_{S}$.
$I_{P} = \frac{V_{S} I_{S}}{V_{P}} = \frac{60}{220} \approx 0.27\, A$.
Thus,the current in the primary coil is $0.27\, A$ and it is a step-down transformer.
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