A charged 10 mu F capacitor is connected to a | vedclass

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

Question

(D) The angular frequency of free oscillations for an $LC$ circuit is given by the formula $\omega_{0} = \frac{1}{\sqrt{LC}}$.Given values are $L = 16

Vedclass · Accepted Answer

$2500 	ext{ rad s}^{-1}$

Vedclass · Answer

(D) The angular frequency of free oscillations for an $LC$ circuit is given by the formula $\omega_{0} = \frac{1}{\sqrt{LC}}$.Given values are $L = 16 	ext{ mH} = 16 	imes 10^{-3} 	ext{ H}$ and $C = 10 \mu F = 10 	imes 10^{-6} 	ext{ F}$.Substituting these values into the formula:$\omega_{0} = \frac{1}{\sqrt{16 	imes 10^{-3} 	imes 10 	imes 10^{-6}}}$$\omega_{0} = \frac{1}{\sqrt{16 	imes 10^{-8}}}$$\omega_{0} = \frac{1}{4 	imes 10^{-4}}$$\omega_{0} = \frac{10^{4}}{4} = 2500 	ext{ rad s}^{-1}$.

$A$ charged $10 \mu F$ capacitor is connected to a $16 \text{ mH}$ inductor. What is the angular frequency of free oscillations of the circuit?

Similar Questions

$A$ $4 \mu F$ 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 $A$)

$A$ transistor-oscillator using a resonant circuit with an inductor $L$ (of negligible resistance) and a capacitor $C$ in series produces oscillations of frequency $f$. If $L$ is doubled and $C$ is changed to $4C$,the frequency will be

$A$ capacitor of capacity $C$ is charged to a potential $V$. It is connected in parallel to an inductor of inductance $L$. The maximum current that will flow in the circuit is

Obtain the differential equation for an $LC$ circuit.

In an $LC$ circuit,the capacitor has maximum charge $q_0$. The value of ${\left| {\frac{{di}}{{dt}}} \right|_{\max }}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module