A transformer operating at primary voltage 8, | vedclass

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Question

(C) For an ideal transformer,the power delivered to the load is the same as the power drawn from the source.Given: Primary voltage $V_p = 8\,kV = 8000

Vedclass · Accepted Answer

$800\,\Omega$ and $0.32\,\Omega$

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(C) For an ideal transformer,the power delivered to the load is the same as the power drawn from the source.Given: Primary voltage $V_p = 8\,kV = 8000\,V$,Secondary voltage $V_s = 160\,V$,Power $P = 80\,kW = 80000\,W$.The resistance in the primary circuit $R_p$ is given by $P = \frac{V_p^2}{R_p}$.$R_p = \frac{V_p^2}{P} = \frac{(8000)^2}{80000} = \frac{64 	imes 10^6}{8 	imes 10^4} = 8 	imes 10^2 = 800\,\Omega$.The resistance in the secondary circuit $R_s$ is given by $P = \frac{V_s^2}{R_s}$.$R_s = \frac{V_s^2}{P} = \frac{(160)^2}{80000} = \frac{25600}{80000} = 0.32\,\Omega$.Thus,the resistances are $800\,\Omega$ and $0.32\,\Omega$.

$A$ transformer operating at primary voltage $8\,kV$ and secondary voltage $160\,V$ serves a load of $80\,kW$. Assuming the transformer to be ideal with a purely resistive load and working on unity power factor,the resistances in the primary and secondary circuits would be:

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