A transformer having efficiency 90 % is worki | vedclass

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Question

(B) Given: Efficiency $\eta = 90 \% = 0.9$, Input Voltage $V_p = 200 \ V$, Input Power $P_{in} = 3 \ kW = 3000 \ W$, Secondary Current $I_s = 6 \ A$.F

Vedclass · Accepted Answer

$450 \ V, 15 \ A$

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(B) Given: Efficiency $\eta = 90 \% = 0.9$, Input Voltage $V_p = 200 \ V$, Input Power $P_{in} = 3 \ kW = 3000 \ W$, Secondary Current $I_s = 6 \ A$.First, calculate the primary current $I_p$ using the input power formula: $P_{in} = V_p 	imes I_p$.$3000 \ W = 200 \ V 	imes I_p \implies I_p = \frac{3000}{200} = 15 \ A$.Next, calculate the output power $P_{out}$ using efficiency: $P_{out} = \eta 	imes P_{in}$.$P_{out} = 0.9 	imes 3000 \ W = 2700 \ W$.Finally, calculate the secondary voltage $V_s$ using the output power formula: $P_{out} = V_s 	imes I_s$.$2700 \ W = V_s 	imes 6 \ A \implies V_s = \frac{2700}{6} = 450 \ V$.Thus, the secondary voltage is $450 \ V$ and the primary current is $15 \ A$.

$A$ transformer having efficiency $90 \%$ is working on $200 \ V$ and $3 \ kW$ power supply. If the current in the secondary coil is $6 \ A$, the voltage across the secondary coil and the current in the primary coil are respectively:

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