A current of 4 text{ A} is flowing at 230 tex | vedclass

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(A) The input power in the primary coil is given by $P_p = V_p 	imes I_p = 230 	ext{ V} 	imes 4 	ext{ A} = 920 	ext{ W}$.Since $40\%$ of the powe

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$0.24$

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(A) The input power in the primary coil is given by $P_p = V_p 	imes I_p = 230 	ext{ V} 	imes 4 	ext{ A} = 920 	ext{ W}$.Since $40\%$ of the power is lost, the efficiency of the transformer is $\eta = 100\% - 40\% = 60\%$.The output power in the secondary coil is $P_s = \eta 	imes P_p = 0.60 	imes 920 	ext{ W} = 552 	ext{ W}$.The output power is also given by $P_s = V_s 	imes I_s$, where $V_s = 2300 	ext{ V}$.Therefore, $I_s = \frac{P_s}{V_s} = \frac{552 	ext{ W}}{2300 	ext{ V}} = 0.24 	ext{ A}$.

$A$ current of $4 \text{ A}$ is flowing at $230 \text{ V}$ in the primary coil of a transformer. If the voltage produced in the secondary coil is $2300 \text{ V}$ and $40\%$ of power is lost, then the current in the secondary will be (in $\text{ A}$)

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