The frequency of oscillation of the current i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) In the given circuit,the two inductors $L$ and $2L$ are connected in series. Therefore,the equivalent inductance is $L_{eq} = L + 2L = 3L$.The two

Vedclass · Accepted Answer

$\frac{1}{6\pi\sqrt{LC}}$

Vedclass · Answer

(B) In the given circuit,the two inductors $L$ and $2L$ are connected in series. Therefore,the equivalent inductance is $L_{eq} = L + 2L = 3L$.The two capacitors $C$ and $2C$ are connected in parallel. Therefore,the equivalent capacitance is $C_{eq} = C + 2C = 3C$.The frequency of oscillation $f$ for an $LC$ circuit is given by the formula $f = \frac{1}{2\pi\sqrt{L_{eq}C_{eq}}}$.Substituting the equivalent values,we get $f = \frac{1}{2\pi\sqrt{(3L)(3C)}} = \frac{1}{2\pi\sqrt{9LC}} = \frac{1}{2\pi 	imes 3\sqrt{LC}} = \frac{1}{6\pi\sqrt{LC}}$.

The frequency of oscillation of the current in the given circuit is:

Similar Questions

$A$ capacitor of capacity '$C$' is charged to a potential difference of '$V_1$'. The plates of the capacitor are then connected to an ideal inductor of inductance '$L$'. The current through the inductor when the potential difference across the capacitor reduces to '$V_2$' is:

$A$ $4 \mu F$ capacitor is charged to $10 \ V$. The battery is then disconnected and a pure $10 \ mH$ coil is connected across the capacitor so that $LC$ oscillations are set up. The maximum current in the coil is (in $A$)

$A$ capacitor of $1\,\mu F$ initially charged to $10\,V$ is connected across an ideal inductor of $0.1\,mH$. The maximum current in the circuit is.....$A$

Explain how $LC$ oscillations take place in the circuit.

$A$ capacitor is fully charged with a battery. Then the battery is removed and a coil is connected with the capacitor in parallel. How does the current vary?

Vedclass Test Series

Exam Paper Generator

Online Exam Module