Physics · 2020
JEE Main · 6 September 2020, Shift 2 · Q23
In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the…
In a series LR circuit, power of $\displaystyle 400$ W is dissipated from a source of $\displaystyle 250 \mathrm{~V}, 50 \mathrm{~Hz}$. The power factor of the circuit is $\displaystyle 0.8$ . In order to bring the power factor to unity, a capacitor of value C is added in series to the L and R. Taking the value of C as $\displaystyle \left(\frac{\mathrm{n}}{3 \pi}\right) \mu \mathrm{F}$, then value of n is \_\_\_\_.
Official answer
400
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JEE Main 2020 Physics question, with the answer from NTA’s final answer key. Where our answers come from.