JEE MAIN - Physics (2022 - 30th June Morning Shift - No. 23)
A series LCR circuit with $$R = {{250} \over {11}}\,\Omega $$ and $${X_L} = {{70} \over {11}}\,\Omega $$ is connected across a 220 V, 50 Hz supply. The value of capacitance needed to maximize the average power of the circuit will be _________ $$\mu$$F. (Take : $$\pi = {{22} \over 7}$$)
Answer
500
Explanation
For maximum power
$$ \begin{aligned} &\text { power factor }=\cos \theta=1\\\\ & \therefore \frac{R}{Z}=1 \\\\ &R^{2}=Z^{2} \\\\ &R^{2}=\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}+\mathrm{R}^{2} \\\\ &\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \\\\ &\frac{70}{11}=\frac{1}{100 \pi \times C} \\\\ &\Rightarrow C=\frac{11}{7000 \pi}=500 \times 10^{-6} F=500 \mu F \end{aligned} $$
$$ \begin{aligned} &\text { power factor }=\cos \theta=1\\\\ & \therefore \frac{R}{Z}=1 \\\\ &R^{2}=Z^{2} \\\\ &R^{2}=\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)^{2}+\mathrm{R}^{2} \\\\ &\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \\\\ &\frac{70}{11}=\frac{1}{100 \pi \times C} \\\\ &\Rightarrow C=\frac{11}{7000 \pi}=500 \times 10^{-6} F=500 \mu F \end{aligned} $$
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