JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 14)
A transformer operating at primary voltage $$8 \,\mathrm{kV}$$ and secondary voltage $$160 \mathrm{~V}$$ serves a load of $$80 \mathrm{~kW}$$. 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
$$800 \,\Omega$$ and $$1.06 \,\Omega$$
$$10 \,\Omega$$ and $$500 \,\Omega$$
$$800 \,\Omega$$ and $$0.32 \,\Omega$$
$$1.06 \,\Omega$$ and $$500 \,\Omega$$
Explanation
$${V_1}{i_1} = {V_2}{i_2} = 80$$ kW
$$ \Rightarrow {i_1} = 10\,A$$ and $${i_2} = {{80 \times 1000} \over {160}} = 500\,A$$
$$ \Rightarrow {R_1} = {{{V_1}} \over {{i_1}}} = 800\,\Omega $$ and $${R_2} = {{160} \over {500}} = 0.32\,\Omega $$
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