Question

An LC circuit contains a 20 mH inductor and a 50 μF capacitor with an initial charge of 10 mC. The resistance of the circuit is negligible. Let the instant the circuit is closed be t = 0. (a) What is the total energy stored initially? Is it conserved during LC oscillations? (b) What is the natural frequency of the circuit? (c) At what time is the energy stored (i) completely electrical (i.e., stored in the capacitor)? (ii) completely magnetic (i.e., stored in the inductor)? (d) At what times is the total energy shared equally between the inductor and the capacitor? (e) If a resistor is inserted in the circuit, how much energy is eventually dissipated as heat?

Answer 1

(a) Total energy is initially in the form of electric field within the plates of charged capacitor.

$U_E=\frac{Q^2}{2C}$

= $\frac{10\times10^{-3})^2}{2\times50\times10^{-6}}$

= 1J

if we neglect the losses due to resistance of connecting wires, the total energy remains conserved during LC oscillations.

(b) Natural frequency of the circuit, f = $\frac{1}{2\pi\sqrt{LC}}$

here, L = 20mH = 20 × 10^-3 H, C = 50μF = 50 × 10^-6 C

so, f = 1/2π√(20 × 10^-3 × 50 × 10^-6)

= 1/2π√(1000 × 10^-9)

= 1/2π√(10^-6)

= 1000/2π = 500/π Hz ≈ 159 Hz.....(1)

(c) instantaneous electrical energy is given by, $U_E=\frac{q_0^2cos^2\omega t}{2C}$

the energy is completely electrical when $\omega t$ = 0, π, 2π, 3π .... or nπ

so, t = $\frac{n\pi}{\omega}=\frac{n\pi}{2\pi f}$ = n/2f

from equation (2), f = 500/π

so, t = nπ/1000sec when , n = 0, 1, 2, 3, 4... or, t = 0, T/2, T , 3T/2, ....

instantaneous magnetic energy,

$U_B=\frac{q_0^2}{2C}sin^2\omega t$

the energy is completely magnetic when $\omega t$ = π/2, 3π/2, 5π/2.... or (2n +1)π/2

then, t = (2n + 1)π/2(2πf) = (2n + 1)/4f

putting f = 500/π , t = (2n + 1)π/1000

where n = 0, 1, 2, 3, ... or t = T/4, 3T/4, 5T/4....

(d) timing for energy shared equally between inductor and capacitor.

$U_B=U_E$

or, $\frac{q_0^2}{2C}sin^2\omega t=\frac{q_0^2}{2C}cos^2\omega t$

or, $tan^2\omega t=1$

or, $tan\omega t=tan\frac{\pi}{4}$

so, t = (2n + 1)π/4$\omega$

where t = π/$\omega$, 3π/4$\omega$, 5π/4$\omega$ or t = T/8, 3T/8, 5T/8 ......

(e) when a resistor is increased in the circuit, eventually all the energy will be lost as heat across resistance . the oscillation will be damped.

Answer 2

Answer:

Here, L=20mH=20×10−3H,C=50μF=50×10−6F,q0=10mC=10×10−3C=10−2C

(a) E=q202C=(10−2)22×50×10−6=1 joule

Yes, this energy is conserved during the oscillations.

(b) v=12πLC−−−√=1×72×2220×10−3×50×10−6−−−−−−−−−−−−−−−−−−√=7×10344=159.1 hertz

(c ) (i) At any instant t, change on the capacitor is q=q0cosωt=q0cos2πTt.

At t=0,T2,T,3T2,......q=q0=max

∴ Energy is stored completely in the capacitor as electrical energy.

(ii) Magnetic energy around L will be maximum, when electrical energy in C is zero, i.e., q = 0

This will be at i=T4,3T4,5T4........

(d) Equal sharing of energy means, energy of capacitor =12 maximum energy.

q22C=12q202C i.e. q=q02–√

From q=q0cosωt=q0cos2πTt

q02–√=q0cos2πTt

cos2πTt=12–√=cos(2n+1)π4∴2πTt=(2n+1)π4

t=(2n+1)T8

Hence, energy will be half on C and half on L at t=T8,3T8,%T8...........

(e) The pressure of resistor in the circuit involves loss of energy. The oscillations become damped and ultimately they disappear, when total energy of 1 joule is dissipated as heat.

Explanation: