In a transformer,the number of turns in prima | vedclass

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(A) For an ideal transformer,the power input equals the power output,which implies $V_{p} i_{p} = V_{s} i_{s}$.Since the voltage ratio is proportional

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$12$

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(A) For an ideal transformer,the power input equals the power output,which implies $V_{p} i_{p} = V_{s} i_{s}$.Since the voltage ratio is proportional to the turns ratio,we have $\frac{V_{s}}{V_{p}} = \frac{N_{s}}{N_{p}} = \frac{i_{p}}{i_{s}}$.Given: $N_{p} = 500$,$N_{s} = 2000$,and $i_{p} = 48 \ A$.Substituting the values into the relation $\frac{N_{s}}{N_{p}} = \frac{i_{p}}{i_{s}}$:$\frac{2000}{500} = \frac{48}{i_{s}}$$4 = \frac{48}{i_{s}}$$i_{s} = \frac{48}{4} = 12 \ A$.

In a transformer,the number of turns in primary and secondary are $500$ and $2000$ respectively. If the current in the primary is $48 \ A$,the current in the secondary is......$A$.

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