A 2,mu F capacitor is charged to 100,V. After | vedclass

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

Question

(C) The energy stored in the capacitor is given by $U_E = \frac{1}{2}CV^2$. When the capacitor discharges through the inductor,this energy is transfer

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) The energy stored in the capacitor is given by $U_E = \frac{1}{2}CV^2$. When the capacitor discharges through the inductor,this energy is transferred to the magnetic field of the inductor,given by $U_B = \frac{1}{2}LI_{\max}^2$. By the law of conservation of energy,$U_E = U_B$,so $\frac{1}{2}CV^2 = \frac{1}{2}LI_{\max}^2$. Substituting the given values: $C = 2 	imes 10^{-6}\,F$,$V = 100\,V$,and $L = 20 	imes 10^{-3}\,H$. $(2 	imes 10^{-6}) 	imes (100)^2 = (20 	imes 10^{-3}) 	imes I_{\max}^2$. $(2 	imes 10^{-6}) 	imes 10^4 = 0.02 	imes I_{\max}^2$. $0.02 = 0.02 	imes I_{\max}^2$. $I_{\max}^2 = 1$,which gives $I_{\max} = 1\,A$.

Similar Questions

The switch is in position $A$ for a long time. At time $t=0$,it is shifted to position $B$. Find the maximum charge that will accumulate on the capacitor.

An $LC$ resonant circuit contains a $400 \, pF$ capacitor and a $100 \, \mu H$ inductor. It is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is

$A$ capacitor of capacitance $C$ has initial charge $Q_0$ and is connected to an inductor of inductance $L$ as shown. At $t=0$,the switch $S$ is closed. Find the current through the inductor when the energy in the capacitor is three times the energy of the inductor.

$A$ tuned circuit of a transistor oscillator unit has an inductance of $5 mH$ and a capacitance of $5 pF$. The natural frequency of the oscillator is

The natural frequency of an $L-C$ circuit is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module