JEE MAIN - Physics (2021 - 27th July Evening Shift - No. 8)
A 100$$\Omega$$ resistance, a 0.1 $$\mu$$F capacitor and an inductor are connected in series across a 250 V supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is 60 Hz.
0.70 H
70.3 mH
7.03 $$\times$$ 10$$-$$5 H
70.3 H
Explanation
C = 0.1 $$\mu$$F = 10$$-$$7 F
Resonant frequency = 60 Hz.
$${\omega _0} = {1 \over {\sqrt {LC} }}$$
$$2\pi {f_0} = {1 \over {\sqrt {LC} }} \Rightarrow L = {1 \over {4{\pi ^2}f_0^2C}}$$
by putting values $$L \simeq 70.3$$ Hz.
Resonant frequency = 60 Hz.
$${\omega _0} = {1 \over {\sqrt {LC} }}$$
$$2\pi {f_0} = {1 \over {\sqrt {LC} }} \Rightarrow L = {1 \over {4{\pi ^2}f_0^2C}}$$
by putting values $$L \simeq 70.3$$ Hz.
Comments (0)
