Name: Purely Inductive Circuit
Uploaded: 2019-07-19
Duration: 5 min 40 s
Description: How does an inductor behave in a circuit? This video will explain every parameter involved in the derivation of a pure inductor working with AC supply.

Question

Using the equation

$\frac{di}{dt} + \frac{R}{L} i + \frac{1}{LC} \int idt = \frac{V_{m}}{L} e^{iωt}$ . for L-C-R circuit, obtain an equation for the complex current $' i'$ and explain its different terms in the equation. Hence obtain the equation for real current $' i'$

Accepted Answer

The given equation is $\frac{di}{dt} + \frac{R}{L} i + \frac{1}{LC} \int idt = \frac{V_{m}}{L} e^{iωt}$

This equation can be written as by multiplying by $\frac{LI}{R}$ as,

$Li \frac{di}{dt} + {Ri}^{2} + \frac{q}{C} i = v i$

Explanation of different terms:

Since $Li \frac{di}{dt} = \frac{d}{dt} (\frac{1}{2} {Li}^{2})$ . =The rate of change of energy stored in an inductor.

${Ri}^{2}$ = joule heating loss.

$\frac{q}{c} i = \frac{d}{dt} (\frac{q^{2}}{2 C})$ =rate of change of energy stored in the capacitor.

$v i = p o w e r$ rate at which driving force pours in energy. It goes into (i) ohmic loss and (ii) increase of stored energy.

The equation can be written as

$L \frac{d^{2} q}{{dt}^{2}} + R \frac{dq}{dt} + \frac{q}{C} = E_{0} sinωt$ . This is like an equation of a forced oscillator.. Its solution is,

$q = q_{o} \sin (ωt + θ) so \frac{dq}{dt} = q_{o} ωcos (ωt + θ) and \frac{d^{2} q}{{dt}^{2}} = - q_{o} ω^{2} \sin (ωt + θ)$

The current in the circuit is given by,

$I = \frac{dq}{dt} = q_{o} ωcos (ωt + ϕ) = q_{o} ωcos (ωt + ϕ - \frac{π}{2})$ , Where $\frac{π}{2}$ is the phase difference

Therefore we can write,

$I = I_{0} \sin (ω t + ϕ)$ .

This is the required euation.

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