Transient response of RL circuit

Figure 6.1: non-realistic model

Figure 6.2: Switch attached to make it realistic

Figure 6.3: Inductance of wire also considered

Figure 6.4: After closing the switch

For DC circuit analysis, the voltage and current source excitation is constant, so C and L are neglected 6.1. The circuit is assumed to be as it is since time=$-\infty$ to $\infty$. In practice, no excitation is constant from $t=-\infty$ to $t=\infty$. A more realistic circuit would include a switch, as shown in Fig.6.2. Also, inductance and capacitances of wires and components cannot be neglected as shown in Fig.6.3, and in Fig.6.4 (for $t\geq 0$). Using KVL:

$$2=i10+L\frac{di}{dt} t>0$$

$$i=0; t\leq 0$$

$$a\frac{dy}{dt}+by=c$$

$$\Rightarrow \frac{dy}{dt}+\frac{b}{a}y=\frac{c}{a}$$

Multiplying both sides by

$$e^{\frac{b}{a}t}$$

to get

$$e^{\frac{b}{a}t}\frac{{\mathrm{d}} y}{{\mathrm{d}} t}+\frac{b}{a}e^{\frac{b}{a}t} y = \frac{c}{a}e^{\frac{b}{a}t}$$

Therefore,

$$\frac{{\mathrm{d}} (ye^{\frac{b}{a}t})}{{\mathrm{d}}t} = \frac{c}{a}e^{\frac{b}{a}t}.$$

Integrating both sides, we get

$$\int\:{\mathrm{d}} (ye^{\frac{b}{a}t})\: = \:\frac{c}{a}\int\:e^{\frac{b}{a}t}\;+\;k.$$

$$\Rightarrow\qquad ye^{\frac{b}{a}t}\; = \;\frac{c}{b}e^{\frac{b}{a}t}\;+\;k.$$

$$\Rightarrow\qquad y\quad = \quad\frac{c}{b}\;+\;k e^{- \frac{b}{a}t}.$$

Note that, at $t\to\infty$, $y\to\frac{c}{b}$ and at $t\to 0$, $y\to\frac{c}{b} + k$.
Now, $L \frac{{\mathrm{d}} i}{{\mathrm{d}} t}\;+\;10 i\;=\;2$, where, $a =\;L$, $b = 10$ and $c = 2$.

$$\Rightarrow\qquad i\;=\;\frac{2}{10} + k e^{- \frac{10}{L} t}.$$

As at $t=0$, $i=0$,

$$\qquad 0\; = \;\frac{2}{10} + k.$$

$$\Rightarrow\qquad k\; = \;- \frac{2}{10}.$$

$$\Rightarrow\qquad i(t)\; = \;\frac{2}{10}\big(1-e^{- \frac{10}{L} t}\big).$$

The plot of $i(t)$ vs. $t$ is shown in the Fig.6.5. Note that when $t\to\infty$, $i(t)\to\frac{2}{10}$A.

Figure 6.5:

Voltage across the inductor is given by $v_L(t)=L\frac{{\mathrm{d}} i}{{\mathrm{d}} t}$. Therefore,

$$v_L(t)\;=\;2 e^{-\frac{10}{L}t}.$$

The plot of $v(t)$ vs $t$ is shown in Fig.6.6.

Figure 6.6:

From the above equation, we notice that in time $\frac{L}{10}$ seconds, the voltage across the inductor would reduce to $\frac{1}{e}$ of its original value and would go on decreasing by a further factor of $\frac{1}{e}$ every $\frac{L}{10}$ seconds thereafter. Therefore, summing it up, we have for an inductor-resistor pair with a constant voltage applied at $t = 0$,

$$i (t) \quad=\quad i_F (1-e^{- \frac{R}{L} t}),$$

$$v (t) \quad=\quad v_I e^{- \frac{R}{L} t}.$$

Figure 6.7:

Now, consider the circuit shown in Fig.6.7.
Before $t = 0$, we have the circuit looking as in Fig.. Therefore we have the initial current (at $t=0$) through the inductor as $0.2$ A.

Figure 6.8:

Figure 6.9:

At $t=0^+$, the circuit looks as in Fig. and therefore, we have the following equations for $t>0$.

$$L\frac{{\mathrm{d}} i}{{\mathrm{d}} t}\;+\;Ri \;=\; 0.$$

$$\Rightarrow\qquad L\frac{{\mathrm{d}} i}{{\mathrm{d}} t} \;=\; -Ri.$$

$$\Rightarrow\qquad \log i \;=\; -\frac{R}{L}t + K$$

$$\Rightarrow\qquad i (t) \quad=\quad K^{'}e^{-\frac{R}{L}t}$$

At $t=0$, $i(t)\;=\;0.2$ A. Hence,

$$i (t)\quad =\quad 0.2e^{-\frac{R}{L}t}.$$

The plot for $i(t)$ vs $t$ would therefore be as in Fig 5.10

Figure 6.10:

Figure 6.11: R $\rightarrow \infty$

Hence, $\frac{L}{R}\approx 0$ and, as shown in 6.11, discharge will be immediate. We write equations for $v_L(t)$ across the inductor.

$$v_L(t)=L\frac{di}{dt}=L\times 0.2 \times \frac{-R}{L}e^{-\frac{R}{L}t}$$

$$=-0.2Re^{-\frac{R}{L}t}$$

Figure 6.12: Note the sign of $v_L(t)$

Sign of $v_L(t)$ is as shown in 6.12. As $R\rightarrow\infty$, $v_L(t=0)\rightarrow\infty$

Figure 6.13: Large inductance doesn't allow currents to change at fast rates

Switching off causes a discharge in the tube or spark at switch 6.13.

Figure 6.14:

At $t=0$, $i=i(0)$ 6.14

$$L\frac{di}{dt}+iR=v$$

$$Li\times e^{\frac{R}{L}t}\vert _{t=0}^{t}=\frac{L}{R}V\times e^{\frac{R}{L}t}\vert _0^t$$

$$i(t)=\frac{V}{R}(1-e^{-\frac{R}{L}t})+i(0)e^{-\frac{R}{L}t}$$

$$=\frac{V}{R}-(\frac{V}{R}-i(0))e^{-\frac{R}{L}t}$$

In generic form, $i(t)=I_F-(I_F-I_I)e^{-\frac{R}{L}t}$

Figure 6.15:

Now, have a look at the circuit shown in the figure 5.15. As the resistance $R$ is 0, the $R-L$ equations are indeterminate and are of the form

$$i(t)=\infty (1-e^{-0.t})=\infty\times 0$$

So, we solve the circuit directly

$$L\frac{{\mathrm{d}} i(t)}{{\mathrm{d}} t} \;=\; 2$$

$$\Rightarrow\qquad i(t) \;=\; \frac{2}{L} t\;+\;k$$

At $t=0$, $i=0$, therefore, $k = 0$

$$\Rightarrow\qquad i(t)\quad = \quad\frac{2}{L} t$$

We will use the above circuit to analyse the circuit shown in Figure 5.16. As the resistance of $5 \Omega$ is in parallel with the voltage source and also the rest of the circuit, the current drawn by it will be constant and will not affect the analysis of the rest of the circuit. So, for $0\leq t <5$, we can consider the circuit to be as in Figure 5.17. Analysing it as in the previous example, we get

$$i(t)\; = \;\frac{10}{10} t$$

$$\Rightarrow\qquad i(t)\; = \;t$$

Figure 6.16:

Further, for $5\leq t <10$, the circuit can be equivalently considered as in Figure 5.19. Notice that still, $i(t)=t$ Amps. as the inductor $10 H$ is in parallel with the voltage source. The plot for $i(t)$ vs. $t$ would therefore be linear as in Figure 5.18. Therefore,

$$i_2 (t)\; = \;1 (1 - e^{ - \frac{10}{5} (t-5)})$$

$$= \;(1 - e^{ -2 (t-5)})$$

$$\Rightarrow \quad i_2 (10)\; = \;10\times0.632$$

$$= \;6.32\;{\mathrm{Amps.}}$$

$$i_1 (t)\; = \;10{\mathrm{Amps.}}$$

Figure 6.17:

After $t > 10$, the circuit can be considered equivalently to be that in Fig 5.20. Now, there is no constant voltage source across the resistance of $5 \Omega$. This, the current flowing through it also comes into the analysis.

Figure 6.18:

$$i_r (t)\; = \;i_1 (t) + i_2 (t)$$

$$\Rightarrow -(i_1 +i_2) 5 = \;-i_r 5\;\;=\;10 \frac{{\mathrm{d}} i_1}{{\mathrm{d}} t}$$

$${\mathrm{and,}}$$

$$- (i_1+i_2) 5\; = \;i_2 + 5\frac{{\mathrm{d}} i_2}{{\mathrm{d}} t}$$

$$\Rightarrow\quad 10 \frac{{\mathrm{d}} i_1}{{\mathrm{d}} t} + 5i_1 + 5i_2\; = \;0$$

$${\mathrm{and,}}\quad 5 \frac{{\mathrm{d}} i_1}{{\mathrm{d}} t} + 5i_1 + 6i_2\; = \;0$$

$$\frac{di_1}{dt}+i_1+i_2=0$$

$$i_1=\frac{di_2}{dt}+\frac{6}{5}i_2$$

$$\Rightarrow 2\frac{d^2i_2}{dt^2}+2\times \frac{6}{5}\frac{di_2}{dt}+\frac{di_2}{dt}+\frac{6}{5}i_2+i_2=0$$

$$2\frac{d^2i_2}{dt^2}+\frac{17}{5}\frac{di_2}{dt}+\frac{11}{5}i_2=0$$

$$To\;solve\;the\;differential\;equation,\;suppose$$

$$\i_2=Ae^{Bt}\;and\;put\;in\;the\;differential\;equation$$

$$We\; get\; B=-0.85\pm i0.063\;(by\;solving\;the\;quadratic\;equation)$$

The solution thus is: $i_2=A_1e^{-0.85 +i0.063}+A_2e^{-0.85 -i0.063}$, where, given the initial conditions, we can solve for $A_1$ and $A_2$.