An ideal transformer has 100 turns in the pri | vedclass

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Question

(A) For an ideal transformer,the ratio of the number of turns is equal to the ratio of the voltages: $\frac{N_S}{N_P} = \frac{V_{S,peak}}{V_{P,peak}}$

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) For an ideal transformer,the ratio of the number of turns is equal to the ratio of the voltages: $\frac{N_S}{N_P} = \frac{V_{S,peak}}{V_{P,peak}}$.Given $N_P = 100$,$N_S = 250$,and $V_{P,peak} = 28\ V$.Substituting the values: $\frac{250}{100} = \frac{V_{S,peak}}{28}$.$V_{S,peak} = 2.5 	imes 28 = 70\ V$.The $r.m.s.$ value of the secondary voltage is given by $V_{S,rms} = \frac{V_{S,peak}}{\sqrt{2}}$.$V_{S,rms} = \frac{70}{1.414} \approx 49.5\ V$.Rounding to the nearest integer,we get $50\ V$.

An ideal transformer has $100$ turns in the primary and $250$ turns in the secondary. The peak value of the $AC$ voltage in the primary is $28\ V$. The $r.m.s.$ secondary voltage is nearest to $....V$.

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