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 _________ $$\mu \mathrm{F}$$.
Respondre
40
Explicació
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
$$ \begin{aligned} & \therefore X_{L}=X_{C} \\\\ & 79.6=\frac{1}{2 \pi(50) \times C} \\\\ & C=\frac{1}{79.6 \times 2 \pi(50)} \\\\ & \approx 40 \mu \mathrm{F} \end{aligned} $$