Response of AC Circuits

Response of AC Circuits

Finding the Response of Series RL Circuit

Consider the following series RL circuit diagram.

Description: Description: Description: RL Series Circuit

In the above circuit, the switch was kept open up to t = 0 and it was closed at t = 0. So, the AC voltage source having a peak voltage of V_m volts is not connected to the series RL circuit up to this instant. Therefore, there is no initial current flows through the inductor. The circuit diagram, when the switch is in closed position, is shown in the following figure.

Description: Description: Description: Closed Series RL Circuit

Now, the current i(t) flows in the entire circuit, since the AC voltage source having a peak voltage of V_m volts is connected to the series RL circuit.

We know that the current i(t) flowing through the above circuit will have two terms, one that represents the transient part and other term represents the steady state.

Mathematically, it can be represented as

$i (t) = i_{T r} (t) + i_{s s} (t)$ (t)=iTr(t)+iss(t)Equation 1

Where,

$i_{T r} (t)$

$K e^{}$ −⟮tτ⟯.

Substitute $i_{T r}$ (t)=Ke−⟮tτ⟯ in Equation 1.

$i (t) = K e^{- ⟮ \frac{t}{τ} ⟯} + i_{s s} (t)$ ⟮tτ⟯+iss(t)Equation 2

Calculation of Steady State Current

If a sinusoidal signal is applied as an input to a Linear electric circuit, then it produces a steady state output, which is also a sinusoidal signal. Both the input and output sinusoidal signals will be having the same frequency, but different amplitudes and phase angles. We can calculate the steady state response of an electric circuit, when it is excited by a sinusoidal voltage source using Laplace Transform approach.

The s-domain circuit diagram, when the switch is in closed position, is shown in the following figure.

Description: Description: Description: Open Series RL Circuit

In the above circuit, all the quantities and parameters are represented in s-domain. These are the Laplace transforms of time-domain quantities and parameters.

The Transfer function of the above circuit is

⇒H(s)=1Z(s)⇒H(s)=1Z(s)

⇒H(s)=1R+sL⇒H(s)=1R+sL

Substitute s=jωs=jω in the above equation.

H(jω)=1R+jωLH(jω)=1R+jωL

Magnitude of H(jω)H(jω) is

|H(jω)|=1R2+ω2−−−−−−−√L2|H(jω)|=1R2+ω2L2

Phase angle of H(jω)H(jω) is

∠H(jω)=−tan−1⟮ωLR⟯∠H(jω)=−tan−1⟮ωLR⟯

We will get the steady state current iss(t)iss(t) by doing the following two steps −

Multiply the peak voltage of input sinusoidal voltage and the magnitude of H(jω)H(jω).

Add the phase angles of input sinusoidal voltage and H(jω)H(jω).

The steady state current iss(t)iss(t) will be

iss(t)=VmR2+ω2L2−−−−−−−−−√sin⟮ωt+φ−tan−1⟮ωLR⟯⟯iss(t)=VmR2+ω2L2sin⟮ωt+φ−tan−1⟮ωLR⟯⟯

Substitute the value of iss(t)iss(t) in Equation 2.

i(t)=Ke−⟮tτ⟯+VmR2+ω2L2√sin⟮ωt+φ−tan−1⟮ωLR⟯⟯i(t)=Ke−⟮tτ⟯+VmR2+ω2L2sin⟮ωt+φ−tan−1⟮ωLR⟯⟯Equation 3

We know that there is no initial current in the circuit. Hence, substitute t = 0& i(t) = 0 in Equation 3 in order to find the value of constant, K.

0=Ke−⟮0τ⟯+VmR2+ω2L2−−−−−−−−−√sin⟮ω(0)+φ−tan−1⟮ωLR⟯⟯0=Ke−⟮0τ⟯+VmR2+ω2L2sin⟮ω(0)+φ−tan−1⟮ωLR⟯⟯

⇒0=K+VmR2+ω2L2−−−−−−−−−√sin⟮φ−tan−1⟮ωLR⟯⟯⇒0=K+VmR2+ω2L2sin⟮φ−tan−1⟮ωLR⟯⟯

⇒K=−VmR2+ω2L2−−−−−−−−−√sin⟮φ−tan−1⟮ωLR⟯⟯⇒K=−VmR2+ω2L2sin⟮φ−tan−1⟮ωLR⟯⟯

Substitute the value of K in Equation 3.

i(t)=−VmR2+ω2L2√sin⟮φ−tan−1⟮ωLR⟯⟯e−⟮tτ⟯+VmR2+ω2L2√sin⟮ωt+φ−tan−1⟮ωLR⟯⟯i(t)=−VmR2+ω2L2sin⟮φ−tan−1⟮ωLR⟯⟯e−⟮tτ⟯+VmR2+ω2L2sin⟮ωt+φ−tan−1⟮ωLR⟯⟯Equation 4

Equation 4 represents the current flowing through the series RL circuit, when it is excited by a sinusoidal voltage source. It is having two terms. The first and second terms represent the transient response and steady state response of the current respectively. We can neglect the first term of Equation 4 because its value will be very much less than one. So, the resultant current flowing through the circuit will be

i(t)=VmR2+ω2L2−−−−−−−−−√sin⟮ωt+φ−tan−1⟮ωLR⟯⟯i(t)=VmR2+ω2L2sin⟮ωt+φ−tan−1⟮ωLR⟯⟯

It contains only the steady state term. Hence, we can find only the steady state response of AC circuits and neglect transient response of it.