A 220, V input is supplied to a transformer. | vedclass

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(D) Given: Input voltage,$V_{p} = 220\, V$. Output voltage,$V_{s} = 440\, V$. Output current,$I_{s} = 2.0\, A$. Efficiency,$\eta = 80\% = 0.8$.The eff

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$5$

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(D) Given: Input voltage,$V_{p} = 220\, V$. Output voltage,$V_{s} = 440\, V$. Output current,$I_{s} = 2.0\, A$. Efficiency,$\eta = 80\% = 0.8$.The efficiency of a transformer is defined as the ratio of output power to input power:$\eta = \frac{P_{out}}{P_{in}} = \frac{V_{s} I_{s}}{V_{p} I_{p}}$Rearranging the formula to solve for the primary current $I_{p}$:$I_{p} = \frac{V_{s} I_{s}}{\eta V_{p}}$Substituting the given values:$I_{p} = \frac{440\, V 	imes 2.0\, A}{0.8 	imes 220\, V}$$I_{p} = \frac{880}{176} = 5\, A$Therefore,the current drawn by the primary windings is $5\, A$.

$A$ $220\, V$ input is supplied to a transformer. The output circuit draws a current of $2.0\, A$ at $440\, V$. If the efficiency of the transformer is $80\%$,the current drawn by the primary windings of the transformer is ..... $A$.

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