A transformer has 100 turns in the primary co | vedclass

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(C) Given: Primary turns $N_P = 100$,Primary current $I_P = 8 	ext{ A}$,Input power $P_i = 1 	ext{ kW} = 1000 	ext{ W}$,Secondary voltage $V_S = 50

Vedclass · Accepted Answer

$400$

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(C) Given: Primary turns $N_P = 100$,Primary current $I_P = 8 	ext{ A}$,Input power $P_i = 1 	ext{ kW} = 1000 	ext{ W}$,Secondary voltage $V_S = 500 	ext{ V}$.Input power $P_i = V_P 	imes I_P$,so $1000 = V_P 	imes 8$.Therefore,primary voltage $V_P = \frac{1000}{8} = 125 	ext{ V}$.Using the transformer ratio formula: $\frac{V_S}{V_P} = \frac{N_S}{N_P}$.Substituting the values: $\frac{500}{125} = \frac{N_S}{100}$.$4 = \frac{N_S}{100}$.$N_S = 400$ turns.

$A$ transformer has $100$ turns in the primary coil and carries $8 A$ current. If the input power is $1 \text{ kW}$,the number of turns in the secondary coil,to have $500 \text{ V}$ output,will be $:-$

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