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(C) Efficiency of a transformer is defined as the ratio of output power to input power: $\eta = \frac{P_{out}}{P_{in}}$.Given, $\eta = 88 \% = 0.88$ a

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$0.45 \,A$

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(C) Efficiency of a transformer is defined as the ratio of output power to input power: $\eta = \frac{P_{out}}{P_{in}}$.Given, $\eta = 88 \% = 0.88$ and $P_{out} = 880 \,W$.Substituting these values: $0.88 = \frac{880}{P_{in}}$.Therefore, $P_{in} = \frac{880}{0.88} = 1000 \,W$.The input power is also given by $P_{in} = V_{in} 	imes I_{in}$, where $V_{in} = 2200 \,V$.Thus, $I_{in} = \frac{P_{in}}{V_{in}} = \frac{1000}{2200} \,A$.$I_{in} = \frac{10}{22} \,A \approx 0.4545 \,A$.Rounding to two decimal places, the input current is $0.45 \,A$.

$A$ step-down transformer reduces the voltage of a transmission line from $2200 \,V$ to $220 \,V$. The power delivered by it is $880 \,W$ and its efficiency is $88 \%$. The input current is

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