A transformer with a turns ratio frac{N_{1}}{ | vedclass

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(A) The turns ratio is given by $\frac{N_{1}}{N_{2}} = \frac{50}{1}$.For an ideal transformer,the voltage ratio is equal to the turns ratio: $\frac{V_

Vedclass · Accepted Answer

$5.76$

Vedclass · Answer

(A) The turns ratio is given by $\frac{N_{1}}{N_{2}} = \frac{50}{1}$.For an ideal transformer,the voltage ratio is equal to the turns ratio: $\frac{V_{1}}{V_{2}} = \frac{N_{1}}{N_{2}}$.Given $V_{1} = 120 \ V$,we have $\frac{120}{V_{2}} = \frac{50}{1}$.Solving for $V_{2}$,we get $V_{2} = \frac{120}{50} = 2.4 \ V$.The power output is dissipated across the secondary resistance $R_{2} = 1 \ \Omega$.Using the formula $P_{	ext{out}} = \frac{V_{2}^{2}}{R_{2}}$,we substitute the values:$P_{	ext{out}} = \frac{(2.4)^{2}}{1} = 5.76 \ W$.

$A$ transformer with a turns ratio $\frac{N_{1}}{N_{2}} = \frac{50}{1}$ is connected to a $120 \ V$ $AC$ supply. If the primary and secondary circuit resistances are $1.5 \ k\Omega$ and $1 \ \Omega$ respectively,find the power output (in $W$).

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