Alternating Current

JEE Main level, single correct, medium

177 of 190

Question 177 of 190

A transformer operating at primary voltage 8kV8kV and secondary voltage 160V160V serves a load of 80kW80kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

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Answer: (C) 800Ω800\Omega and 0.32Ω0.32\Omega

Since the transformer is assumed to be purely resistive, the power of the transformer can be calculated with P=V2RP=\dfrac{{V}^{2}}{R}. Therefore,
For the primary coil,
P=(8×103)2RP=80×103WP=\dfrac{{\left(8\times {10}^{3}\right)}^{2}}{{R}_{P}}=80\times {10}^{3}W
⇒RP=800Ω\Rightarrow {R}_{P}=800\Omega
For the secondary coil,
P=(160)2RS=80×103WP=\dfrac{{\left(160\right)}^{2}}{{R}_{S}}=80\times {10}^{3}W
⇒RS=0.32Ω\Rightarrow {R}_{S}=0.32\Omega

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