To find the step response of an RL circuit, the following is required:
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The initial inductor current \(i(0)\).
Since the inductor current cannot change instantaneously,
\(\boxed{i\left(0^-\right) = i\left(0^+\right) = i\left(0\right) = I_0}\)
where \(i\left(0^-\right)\) is the current through the inductor just before switching and \(i\left(0^+\right)\) is its current immediately after switching.
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The final (steady state) inductor current \(i(\infty)\).
Use the fact that the inductor acts like a short circuit to DC at steady state.
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The time constant \(\tau\).
The time constant \(\tau\) for an RL circuit is
\(\boxed{\tau = \frac{L}{R}}\)
where \(R\) is the thevenin resistance at the inductor terminals and \(L\) is the equivalent inductance.
When the DC source of an RL circuit is suddenly applied, the voltage or current source can be modeled as a step function, and the response is known as a step response.
Let the response be the sum of the transient response and the steady-state response,
\(\displaystyle i = i_t + i_{ss}\)
The transient response \(i_t\) is always a decaying exponential,
\(\displaystyle i_t = Ae^{\displaystyle -t/\tau}, \quad \tau = \frac{L}{R}\)
where \(A\) is the constant to be determined.
The steady-state response \(i_{ss}\) is the portion of complete response \(i\) that remains after the transient response \(i_t\) has died out.
\(\displaystyle i_{ss} = \frac{V_s}{R}\)
Determine constant \(A\) using the initial value \(I_0\)
\(\displaystyle i\left(t\right) = i_t + i_{ss} = Ae^{\displaystyle -t/\tau} + \frac{V_s}{R}\)
\(\displaystyle i\left(0\right) = I_0 = A + \frac{V_s}{R}\)
\(\displaystyle A = I_0 - \frac{V_s}{R}\)
Substitute \(A\) to \(i\left(t\right)\)
\(\boxed{i\left(t\right) = \frac{V_s}{R} + \left(I_0 - \frac{V_s}{R}\right) e^{\displaystyle -t/\tau}}\)
where time constant \(\displaystyle \tau = \frac{L}{R}\)
This is known as the complete response of the RL circuit to a sudden application of a DC source.
The complete response may be written as
\(\boxed{i\left(t\right) = i\left(\infty\right) + \left[i\left(0\right) - i\left(\infty\right)\right]\,e^{\displaystyle -t/\tau}, \quad t > 0}\)
where time constant \(\displaystyle \tau = \frac{L}{R}\), \(i\left(0\right)\) is the initial voltage at \(t = 0^+\) and \(i\left(\infty\right)\) is the final or steady-state value.
This equation can be used in source-free RL circuit.
If the switch changes position at time \(t = t_0\) instead of at \(t = 0\), there is a time delay in the response.
\(\boxed{i\left(t\right) = i\left(\infty\right) + \left[i\left(t_0\right) - i\left(\infty\right)\right]\,e^{\displaystyle -\left(t - t_0\right)/\tau}, \quad t > t_0}\)
where time constant \(\displaystyle \tau = \frac{L}{R}\) and \(i\left(t_0\right)\) is the initial value at \(t = t_{0}^{+}\).
This equation can be used in source-free RL circuit.