JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 25)
A series LCR circuit is connected to an ac source of $$220 \mathrm{~V}, 50 \mathrm{~Hz}$$. The circuit contain a resistance $$\mathrm{R}=100 ~\Omega$$ and an inductor of inductive reactance $$\mathrm{X}_{\mathrm{L}}=79.6 ~\Omega$$. The capacitance of the capacitor needed to maximize the average rate at which energy is supplied will be _________ $$\mu \mathrm{F}$$.
답변
40
설명
To maximize the average rate at which energy
supplied i.e. power will be maximum.
So in LCR circuit power will be maximum at the condition of resonance and in resonance condition
$$ \begin{aligned} & \therefore X_{L}=X_{C} \\\\ & 79.6=\frac{1}{2 \pi(50) \times C} \\\\ & C=\frac{1}{79.6 \times 2 \pi(50)} \\\\ & \approx 40 \mu \mathrm{F} \end{aligned} $$
So in LCR circuit power will be maximum at the condition of resonance and in resonance condition
$$ \begin{aligned} & \therefore X_{L}=X_{C} \\\\ & 79.6=\frac{1}{2 \pi(50) \times C} \\\\ & C=\frac{1}{79.6 \times 2 \pi(50)} \\\\ & \approx 40 \mu \mathrm{F} \end{aligned} $$