A current of 5 text{ A} is flowing at 220 tex | vedclass

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(C) The input power in the primary coil is given by $P_{in} = V_p 	imes I_p = 220 	ext{ V} 	imes 5 	ext{ A} = 1100 	ext{ W}$.Given that $50 \%$ o

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$0.25$

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(C) The input power in the primary coil is given by $P_{in} = V_p 	imes I_p = 220 	ext{ V} 	imes 5 	ext{ A} = 1100 	ext{ W}$.Given that $50 \%$ of the power is lost, the output power in the secondary coil is $P_{out} = 50 \% 	ext{ of } P_{in} = 0.5 	imes 1100 	ext{ W} = 550 	ext{ W}$.The output power is also given by $P_{out} = V_s 	imes I_s$, where $V_s = 2200 	ext{ V}$.Substituting the values, $550 	ext{ W} = 2200 	ext{ V} 	imes I_s$.Therefore, $I_s = \frac{550}{2200} 	ext{ A} = 0.25 	ext{ A}$.

$A$ current of $5 \text{ A}$ is flowing at $220 \text{ V}$ in a primary coil of a transformer. If the voltage produced in the secondary coil is $2200 \text{ V}$ and $50 \%$ of power is lost, then the current in the secondary coil will be (in $\text{ A}$)

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