Alternating Current Concept Page - 5

Definition

Similarity between electrical and mechanical oscillations

The LC oscillation is similar to the mechanical oscillation of a blocka ttached to a spring. For a block of mass m$m$ oscillating with frequency ω0${\omega }_0$, the equation is:
d2xdt2+ω20x=0${\displaystyle \frac{d^{2}x}{d{t}^2}}+{\omega }_0^{2}x=0$

Here, ω0=km${\omega }_0={\displaystyle \frac{k}{m}}$, and k$k$ is the spring constant.
 So, x$x$ corresponds to q$q$.
 In case of a mechanical system F=ma=m(dvdt)=m(d2xdt2)$F=ma=m({\displaystyle \frac{dv}{dt}})=m({\displaystyle \frac{d^{2}x}{d{t}^2}})$.
 For an electrical system, ϵ=L(didt)=L(d2qdt2)$ϵ=L({\displaystyle \frac{di}{dt}})=L({\displaystyle \frac{d^{2}q}{d{t}^2}})$.
Comparing these two equations, we see that L$L$ is analogous to mass m$m$: L$L$ is a measure of resistance to change in current. In case of LC circuit, ω0=1LC${\omega }_0={\displaystyle \frac{1}{LC}}$ and for mass on a spring,ω0=km${\omega }_0={\displaystyle \frac{k}{m}}$.
 So, 1C${\displaystyle \frac{1}{C}}$ is analogous to k$k$. The constant k$k$ (=F/x)$(=F/x)$ tells us the (external) force required to produce a unit displacement whereas 1/C$1/C$ (=V/q)$(=V/q)$ tells us about the potential difference required to store a unit charge.

Formula

Natural frequency in LC circuit

The natural frequency of oscillations ω0${\omega }_0$ in an LC circuit is given as 
ω0=1LC−−−√${\omega }_0={\displaystyle \frac{1}{\sqrt{LC}}}$

Formula

Charge variation in an LC circuit

q=qmcos(ω0t+ϕ)$q=q_{m}cos({\omega }_{0}t+\varphi )$ where 
q$q$ is the charge in the circuit at time t,
qm$q_m$ is the charge in the circuit at time t=0,
ω0=1L√C${\omega }_0={\displaystyle \frac{1}{\sqrt{L}C}}$ is the natural frequency,
ϕ$\varphi$ is the phase constant.

Example

Charge Across Inductor/Capacitor in a LC circuit

Example: 
In an oscillating LC circuit the maximum charge on the  capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is:Solution:

For LC circuit, UE+UB=Q22C$U_E+U_B={\displaystyle \frac{Q^2}{2C}}$

Here UE=UB$U_E=U_B$ and UE=q22C$U_E={\displaystyle \frac{q^2}{2C}}$   where q$q$ is the required charge on the capacitor.

So, 2UE=Q22C$2U_E={\displaystyle \frac{Q^2}{2C}}$

or 2×q22C=Q22C$2\times {\displaystyle \frac{q^2}{2C}}={\displaystyle \frac{Q^2}{2C}}$

or q=Q2√$q={\displaystyle \frac{Q}{\sqrt{2}}}$

Diagram

Find energy stored across inductor/capacitor at a given time instant in a LC circuit

The LC oscillation is similar to the mechanical oscillation of a block attached to a spring. The lower part of each figure in the diagram depicts the corresponding stage of a mechanical system (a block attached to a spring).

Example

Problem on Energy in LC oscillation

Example: In an oscillating LC circuit the maximum charge on the capacitor is Q.   Find the charge on the capacitor when the energy is stored equally between the electric and magnetic field.

Solution:
For LC circuit, UE+UB=Q22C$U_E+U_B={\displaystyle \frac{Q^2}{2C}}$
here UE=UB$U_E=U_B$ and UE=q22C$U_E={\displaystyle \frac{q^2}{2C}}$
  where q is the required charge on the capacitor.
So, 2UE=Q22C$2U_E={\displaystyle \frac{Q^2}{2C}}$or
 2×q22C=Q22C$2\times {\displaystyle \frac{q^2}{2C}}={\displaystyle \frac{Q^2}{2C}}$or
q=Q2√$q={\displaystyle \frac{Q}{\sqrt{2}}}$

Example

Basic problems on LC oscillations

Example: An LC circuit contains a 20 mH$20mH$ inductor and a 50$50$ μ$\mu$F capacitor with an initial charge of 10 mC$10mC$. The resistance of the circuit is negligible.Let the instant the circuit is closed be t=0$t=0$.
(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?

Solution:
(a)The energy stored initially is E=Q22C=(10×10−3)22×50×10−6=1J$E={\displaystyle \frac{Q^2}{2C}}={\displaystyle \frac{(10\times {10}^{-3}{)}^2}{2\times 50\times {10}^{-6}}}=1J$
Yes. The energy is conserved. It only gets converted from electrical to magnetic and vice versa.
(b) All units in SI. Natural frequency of the circuit is given by,
ν=1/2πLC−−−√$\nu =1/2\pi \sqrt{LC}$

ν=159.24Hz$\nu =159.24Hz$

(c)
(i)Time period, T=1/f=6.28 ms$T=1/f=6.28ms$
total charge on capacitor at time t=Q′=Qcos(2πTt)$t=Q^{{}^{\prime }}=Qcos\left({\displaystyle \frac{2\pi }{T}}t\right)$
For energy stored in electrical, we can write, Q′=±Q$Q^{{}^{\prime }}=\pm Q$
It can be inferred that the energy stored in capacitor is completely electrical at time, t=0,T/2,3T/2,...$t=0,T/2,3T/2,...$
(ii) Magnetic energy is maximum when the electrical energy is zero. hence, t=T/4,3T/4,5T/4,...$t=T/4,3T/4,5T/4,...$
(d) Q1$Q_1$is the charge when total energy is equally shared.
⇒12Q21C=12(12Q2C)$\Rightarrow {\displaystyle \frac{1}{2}}{\displaystyle \frac{Q_1^2}{C}}={\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{2}}{\displaystyle \frac{Q^2}{C}})$Q1=±Q/2√$Q_1=\pm Q/\sqrt{2}$
but, Q1=Qcos2πt/T$Q_1=Qcos2\pi t/T$
⇒cos2πt/T=±1/2√⇒t=(2n+1)T/8$\Rightarrow cos2\pi t/T=\pm 1/\sqrt{2}\Rightarrow t=(2n+1)T/8$
so, t=T/8, 3T/8, 5T/8$t=T/8,3T/8,5T/8$(e)The energy stored initially is E=Q22C=(10×10−3)22×50×10−6=1J$E={\displaystyle \frac{Q^2}{2C}}={\displaystyle \frac{(10\times {10}^{-3}{)}^2}{2\times 50\times {10}^{-6}}}=1J$
When the resistor is inserted in the circuit, it will dissipate entire 1 J$1J$ in heat.

Definition

Analogies between mechanical and electrical oscillations

Mechanical oscillations	Electrical oscillations
Mass m	Inductance L
Force constant k	Reciprocal capacitance 1C$\frac{1}{C}$
Displacement x	Charge q
Velocity v=dx/dt	Current i=dq/dt
Mechanical energy	Electrical energy
E=12kx2+12mv2$E=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2$	U=q2C+12Li2$U=\frac{q^2}{C}+\frac{1}{2}L{i}^2$

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