Table of contents | |
Transient Response | |
Presence or Absence of Transients | |
Inductor Behavior | |
Capacitor Behavior | |
Steady state Response |
Assume the switching action takes place at t = 0. Inductor current does not change instantaneously, when the switching action takes place. That means, the value of inductor current just after the switching action will be same as that of just before the switching action.
Mathematically, it can be represented as
iL(0+)=iL(0−)
The capacitor voltage does not change instantaneously similar to the inductor current, when the switching action takes place. That means, the value of capacitor voltage just after the switching action will be same as that of just before the switching action.
Mathematically, it can be represented as
vc(0+) = vc(0−)
The part of the time response that remains even after the transient response has become zero value for large values of ‘t’ is known as steady state response. This means, there won’t be any transient part in the response during steady state.
If the independent source is connected to the electric circuit or network having one or more inductors and resistors (optional) for a long time, then that electric circuit or network is said to be in steady state. Therefore, the energy stored in the inductor(s) of that electric circuit is of maximum and constant.
Mathematically, it can be represented as
= Maximum & constant
⇒ iL = Maximum & constant
Therefore, inductor acts as a constant current source in steady state.
The voltage across inductor will be
So, the inductor acts as a short circuit in steady state.
If the independent source is connected to the electric circuit or network having one or more capacitors and resistors (optional) for a long time, then that electric circuit or network is said to be in steady state. Therefore, the energy stored in the capacitor(s) of that electric circuit is of maximum and constant.
Mathematically, it can be represented as
Maximum & constant
⇒ vc = Maximum & constant
Therefore, capacitor acts as a constant voltage source in steady state.
The current flowing through the capacitor will be
So, the capacitor acts as an open circuit in steady state.
Consider the following series RL circuit diagram.
In the above circuit, the switch was kept open up to t = 0 and it was closed at t = 0. So, the DC voltage source having V volts is not connected to the series RL circuit up to this instant. Therefore, there is no initial current flows through inductor.
The circuit diagram, when the switch is in closed position is shown in the following figure.
Now, the current i flows in the entire circuit, since the DC voltage source having V volts is connected to the series RL circuit.
Now, apply KVL around the loop.
Equation 1
The above equation is a first order differential equation and it is in the form of
Equation 2
By comparing Equation 1 and Equation 2, we will get the following relations.
x = t
y = i
P = R/L
Q = V/L
The solution of Equation 2 will be
Equation 3
Where, k is the constant.
Substitute, the values of x, y, P & Q in Equation 3.
Equation 4
We know that there is no initial current in the circuit. Hence, substitute, t = 0 and 𝑖 = 0 in Equation 4 in order to find the value of the constant k.
Substitute, the value of k in Equation 4.
Therefore, the current flowing through the circuit is
Equation 5
So, the response of the series RL circuit, when it is excited by a DC voltage source, has the following two terms.
We can re-write the Equation 5 as follows −
Equation 6
Where, τ is the time constant and its value is equal to L/R.
Both Equation 5 and Equation 6 are same. But, we can easily understand the above waveform of current flowing through the circuit from Equation 6 by substituting a few values of t like 0, τ, 2τ, 5τ, etc.
In the above waveform of current flowing through the circuit, the transient response will present up to five time constants from zero, whereas the steady state response will present from five time constants onwards.
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