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(C) For an ideal transformer, the power input in the primary coil is equal to the power output in the secondary coil.Given: Power output $(P_{out})$ =

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$0.36$

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(C) For an ideal transformer, the power input in the primary coil is equal to the power output in the secondary coil.Given: Power output $(P_{out})$ = $80 	ext{ W}$, Primary voltage $(V_p)$ = $220 	ext{ V}$.Since the transformer is ideal, $P_{in} = P_{out} = 80 	ext{ W}$.The power input is given by the formula $P_{in} = V_p 	imes I_p$, where $I_p$ is the current in the primary winding.Therefore, $I_p = \frac{P_{in}}{V_p} = \frac{80 	ext{ W}}{220 	ext{ V}}$.$I_p = \frac{8}{22} 	ext{ A} \approx 0.3636 	ext{ A}$.Rounding to two decimal places, the current in the primary winding is approximately $0.36 	ext{ A}$.

An $80 \text{ W}$ lamp is connected to the secondary of a step-down transformer, where the primary is connected to an $AC$ mains of $220 \text{ V}$. Assuming the transformer to be ideal, the current in the primary winding is nearly: (in $\text{ A}$)

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