To find the natural 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 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.
A source-free RL circuit occurs when its DC source is suddenly disconnected. The energy already stored in the inductor is released to the resistors.
Assume \(i\left(t\right)\) is the current through the inductor. At \(t = 0\), the inductor has an initial current \(I_0\)
\(\boxed{i\left(0\right) = I_0}\)
with the corresponding energy stored in the inductor as
\(\boxed{w\left(0\right) = \frac{1}{2} LI_{0}^2}\)
Applying KVL around the loop
\(\displaystyle v_L + v_R = 0\)
\(\displaystyle L\frac{di}{dt} + Ri = 0\)
\(\displaystyle \frac{di}{dt} + \frac{R}{L}i = 0\)
\(\displaystyle \frac{di}{i} = -\frac{R}{L}dt\)
Integrating gives
\(\displaystyle \int_{I_0}^{i\left(t\right)}\frac{di}{i} = -\int_{0}^{t}\frac{R}{L}dt\)
\(\displaystyle \left.\ln{\left(i\right)}\right|_{I_0}^{i\left(t\right)} = -\left.\frac{Rt}{L}\right|_{0}^{t}\)
\(\displaystyle \ln{\left[i\left(t\right)\right]} - \ln{I_0} = -\frac{Rt}{L}\)
\(\displaystyle \ln{\displaystyle \frac{i\left(t\right)}{I_0}} = -\frac{Rt}{L}\)
\(\displaystyle \frac{i\left(t\right)}{I_0} = e^{\displaystyle -\frac{Rt}{L}}\)
\(\boxed{i\left(t\right) = I_0\,e^{\displaystyle -\frac{Rt}{L}} = I_0\,e^{\displaystyle -\frac{t}{\tau}}}\)
where time constant \(\displaystyle \tau = \frac{L}{R}\)
This shows that the natural response of the RL circuit is an exponential decay of the initial current \(I_0\).
The energy that was initially stored in the inductor is eventually dissipated in the resistor.
The power dissipated in the resistor is
\(\displaystyle v_R \left(t\right) = iR = I_{0}R e^{\displaystyle -t/\tau}\)
\(\boxed{p\left(t\right) = v_{R}i = I_{0}^{2}R e^{\displaystyle -2t/\tau}}\)
The energy absorbed by the resistor up to time \(t\) is
\(\displaystyle w_{R}\left(t\right) = \int_{0}^{t} p\left(\lambda\right)\,d\lambda = \int_{0}^{t} I_{0}^{2}R e^{\displaystyle -2\lambda/\tau}\,d\lambda\)
\(\displaystyle w_{R}\left(t\right) = -\left.\frac{\tau}{2} I_{0}^{2}R e^{\displaystyle -2\lambda/\tau}\right|_{0}^{t},\quad \tau = \frac{L}{R}\)
\(\boxed{w_{R}\left(t\right) = \frac{1}{2} LI_{0}^{2} \left(1 - e^{\displaystyle -2t/\tau}\right)}\)
where time constant \(\displaystyle \tau = \frac{L}{R}\)