In a series LR circuit, power… - JEE Main 2020 Physics PYQ

JEE Main 2020 — Physics Electromagnetism

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In a series $\mathrm{LR}$ circuit, power of $400W$ is dissipated from a source of $250V,50\mathrm{Hz}$ . The power factor of the circuit is $0.8$ . 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 $(\frac{n}{3\pi })\mu F$ , then value of $n$ is

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Held on 6 Sept 2020 · Verified 6 Jul 2026.

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Solution

$P={V}_{m}$ . in $cos\phi$

$400=250\times 1m\times 0.8$

${i}_{\mathrm{rms}}=2A$

${(1m)}^{2}.R=P$

$4\times R=400$

$\Rightarrow$ $R=100\Omega$ .

$cos\phi =\frac{R}{\sqrt{{R}^{2}+{X}_{L}^{2}}}$

${100}^{2}+{X}_{L}^{2}={(\frac{100}{0.8})}^{2}$

${X}_{L}=75\Omega$

Power factor is unity

${X}_{C}={X}_{L}=75$

$\frac{1}{\omega }=75$

$\Rightarrow$ $C=\frac{1}{75\times 2H\times 50}=\frac{1}{7500\pi }F$

$3\pi \times 2500$

$=\frac{1}{3\pi }\times 4\times {10}^{2} \mathrm{mF}$

$=\frac{400}{3\pi }\mu F$

$N=400$

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Solution

$P={V}_{m}$ . in $cos\phi$

$400=250\times 1m\times 0.8$

${i}_{\mathrm{rms}}=2A$

${(1m)}^{2}.R=P$

$4\times R=400$

$\Rightarrow$ $R=100\Omega$ .

$cos\phi =\frac{R}{\sqrt{{R}^{2}+{X}_{L}^{2}}}$

${100}^{2}+{X}_{L}^{2}={(\frac{100}{0.8})}^{2}$

${X}_{L}=75\Omega$

Power factor is unity

${X}_{C}={X}_{L}=75$

$\frac{1}{\omega }=75$

$\Rightarrow$ $C=\frac{1}{75\times 2H\times 50}=\frac{1}{7500\pi }F$

$3\pi \times 2500$

$=\frac{1}{3\pi }\times 4\times {10}^{2} \mathrm{mF}$

$=\frac{400}{3\pi }\mu F$

$N=400$

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Solution

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