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

Write the differential equation for charge in L-C-R series A.C. circuit and obtain an equation for complex current from it.

Accepted Answer

Consider the following circuit having a resistor $R$ , an inductor $L$ , and a capacitor $C$ , connected in series, having an AC voltage source $V$ , as the voltage is sinusoidal in nature hence, $V = V_{m} \sin ω t$ , where $V_{m}$ is the amplitude of the voltage and $ω$ is the frequency.

If $i$ is the current at time $t$ , and $q$ is the charge on capacitor, then from Krichoff's loop rule

$L \frac{d i}{d t} + i R + \frac{q}{C} = V - - - - - - - - - (1)$ .

For an instantaneous current at time $t$ , $i = \frac{d q}{d t}$ '

And, $\frac{d i}{d t} = \frac{d^{2} q}{d t^{2}}$

Putting this value of instantaneous current in equation $(1)$

$L \frac{d^{2} q}{d t^{2}} + R \frac{d q}{d t} + \frac{q}{C} = V_{m} \sin ω t$

The above equation can be considered analogous to the damped oscillator equation. In order to solve this equation let us consider,

$q = q_{m} \sin (ω t + θ)$

$\frac{d q}{d t} = q_{m} ω \cos (ω t + θ)$

and further, $\frac{d^{2} q}{d t^{2}} = - q_{m} ω^{2} \sin (ω t + θ)$

Now, substitute the obtained values in the voltage equation

As, $X_{C} = \frac{1}{ω C} and X_{L} = ωL$

$q_{m} ω [R \cos (ω t + θ) + (X_{C} - X_{L}) \sin (ω t + θ) = V_{m} \sin ω t$

As we know that impedance $Z = \sqrt{R^{2} + {(X_{c} - X_{L})}^{2}}$

Therefore, if we substitute the values in the above expression of impedance, we get

$\frac{q_{m} ω}{Z} [\frac{R}{Z} \cos (ω t + θ) + \frac{(X_{C} - X_{L})}{Z} \sin (ω t + θ)] = V_{m} \sin ω t$

Now, let $\frac{R}{Z} = \cos ϕ$ ,and $\frac{(X_{C} - X_{L})}{Z} = \sin ϕ$ .

Then, $ϕ = \tan^{- 1} \frac{(X_{C} - X_{L})}{R}$ ,

$q_{m} ω Z [\cos (ω t + θ - ϕ)] = V_{m} \sin ω t$

Now, comparing this equation we get, $q_{m} ω Z = V_{m} = i_{m} Z$ .

And, $θ - ϕ = - \frac{π}{2}$

$θ = - \frac{π}{2} + ϕ$

Hence, the equation for current in the circuit can be given as,

$i = \frac{d q}{d t} = q_{m} ω \cos (ω t + θ) = i_{m} \cos (ω t + θ) = i_{m} \sin (ω t + θ)$ .

Write the differential equation for charge in L-C-R series A.C. circuit and obtain an equation for complex current from it.

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