A current of 5 ~A is flowing at 220 ~V in the | vedclass

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(C) Given: Primary voltage $V_{p} = 220 ~V$,Primary current $I_{p} = 5 ~A$,Secondary voltage $V_{s} = 2200 ~V$,Power loss $= 50 \%$.Efficiency $\eta =

Vedclass · Accepted Answer

$0.25$

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(C) Given: Primary voltage $V_{p} = 220 ~V$,Primary current $I_{p} = 5 ~A$,Secondary voltage $V_{s} = 2200 ~V$,Power loss $= 50 \%$.Efficiency $\eta = 100 \% - 50 \% = 50 \%$.The efficiency of a transformer is defined as the ratio of output power to input power:$\eta = \frac{P_{out}}{P_{in}} = \frac{V_{s} I_{s}}{V_{p} I_{p}}$Substituting the given values:$0.5 = \frac{2200 	imes I_{s}}{220 	imes 5}$$0.5 = \frac{2200 	imes I_{s}}{1100}$$0.5 = 2 	imes I_{s}$$I_{s} = \frac{0.5}{2} = 0.25 ~A$.Therefore,the current in the secondary coil is $0.25 ~A$.

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

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