Chapter 5, Problem 5.69HP

Assume the switch in the circuit in Figure P5.69has been closed for a very long lime. It is suddenly opened at $t=0$ and then reclosed at $t=5s$ .Determine the inductor current $i_L$ and the voltage v across the $2\text{-}\Omega$ resistor for $t\ge 0$.

To determine

The inductor current $i_L$ and voltage $v$ across the 2Ω resister for $t\ge 0$ .

Answer to Problem 5.69HP

  $\begin{array}{l}{i}_{Lo}\left(t\right)=\left[\begin{array}{l}2.1+e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)+\\ e^{-0.042\left(t-5\right)}\left(-3.72\cos \left(0.220\left(t-5\right)\right)+1.8\sin \left(0.220\left(t-5\right)\right)\right)\text{A}\end{array}\right]\\ \\ V_{2\Omega }=6+2e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)\end{array}$

Explanation of Solution

Given:

The given circuit is shown below:

  

Calculation:

When switch is closed for long time means the circuit is in steady state condition. In this condition the capacitor is open circuited, and inductor is short circuited. Therefore, the modified circuit can be drawn as:

  

The current through the source just before the opening it at t = 0,

  $\begin{array}{l}i=\frac{6}{3\parallel 2}\\ i=\frac{6}{\frac{3\times 2}{3+2}}\\ i=\frac{6}{6}\times 5\\ i=5\text{A}\end{array}$

Applying the current division rule, the initial value of the current can be calculated as

  $\begin{array}{l}{i}_L\left(0^{+}\right)=\left(\frac{2}{3+2}\right)5\\ i_L\left(0^{+}\right)=2\text{A}\end{array}$

The initial value of voltage across the capacitor will be

  $\begin{array}{l}{v}_C\left(0^{+}\right)=3i_L\left(0^{+}\right)\\ v_C\left(0^{+}\right)=3\left(2\right)\\ v_C\left(0^{+}\right)=6\text{V}\end{array}$

The switch is opened at t = 0, the circuit can be drawn as

  

Applying KVL in the loop having current $I_1$ ,

  $\begin{array}{l}2I_1+3\left(I_1-I_2\right)+5\frac{d{I}_1}{dt}=0\\ 5I_1-3I_2+5\frac{d{I}_1}{dt}=0\\ I_2=\frac{1}{3}\left(5I_1+5\frac{d{I}_1}{dt}\right)\end{array}$

Applying KVL in the loop having current $I_2$ ,

  $3\left(I_2-I_1\right)+\frac{1}{4}{\displaystyle \int I_{2}dt}=0$

Differentiating with respect to t,

  $\begin{array}{l}\frac{d}{dt}\left(3\left(I_2-I_1\right)+\frac{1}{4}{\displaystyle \int I_{2}dt}\right)=0\\ 3\frac{d{I}_2}{dt}-3\frac{d{I}_1}{dt}+\frac{1}{4}{I}_2=0\end{array}$

Put the value of $I_2$ ,

  $\begin{array}{l}3\frac{d}{dt}\left(\frac{1}{3}\left(5I_1+5\frac{d{I}_1}{dt}\right)\right)-3\frac{d{I}_1}{dt}+\frac{1}{4}\left(\frac{1}{3}\left(5I_1+5\frac{d{I}_1}{dt}\right)\right)=0\\ 5\frac{d{I}_1}{dt}+5\frac{d^2{I}_1}{d{t}^2}-3\frac{d{I}_1}{dt}+\frac{5}{12}{I}_1+\frac{5}{12}\frac{d{I}_1}{dt}=0\\ 5\frac{d^2{I}_1}{d{t}^2}+\frac{29}{12}\frac{d{I}_1}{dt}+\frac{5}{12}{I}_1=0\\ \frac{d^2{I}_1}{d{t}^2}+\frac{29}{60}\frac{d{I}_1}{dt}+\frac{5}{60}{I}_1=\mathrm{0...........}\left(1\right)\end{array}$

Assuming $\frac{d}{dt}=D$,

  $\begin{array}{l}{D}^2{I}_1+\frac{29}{12}D{I}_1+\frac{5}{12}{I}_1=0\\ \left(D^2+\frac{29}{12}D+\frac{5}{12}\right)I_1=0\end{array}$

The auxiliary equation can be written as

  $D^2+\frac{29}{12}D+\frac{5}{12}=0$

The roots can be calculated as

  $\begin{array}{l}D=\frac{-\frac{29}{12}\pm \sqrt{{\left(\frac{29}{12}\right)}^2-\left(4\right)\left(1\right)\left(\frac{5}{12}\right)}}{2\left(1\right)}\\ D=-0.242+j0.159\end{array}$

Therefore, the solution of the differential equation will be in the form

  $I_1\left(t\right)=e^{-0.242t}\left(A\cos \left(0.159t\right)+B\sin \left(0.159t\right)\right)$

Current through the inductor is equal to the loop current,

  $i_L\left(t\right)=e^{-0.242t}\left(A\cos \left(0.159t\right)+B\sin \left(0.159t\right)\right)\mathrm{....................}\left(2\right)$

Using the initial conditions,

At t = 0,

  $\begin{array}{l}{i}_L\left(0\right)=i\left(0^{+}\right)\\ i_L\left(0\right)=2\text{A}\end{array}$

Substituting 0 for t,

  $i_L\left(0\right)=A$

Therefore,

  $A=2$

Differentiating equation 2 with respect to t,

  $\begin{array}{l}\frac{d}{dt}{i}_L\left(t\right)=\frac{d}{dt}\left[e^{-0.242t}\left(A\cos \left(0.159t\right)+B\sin \left(0.159t\right)\right)\right]\\ \frac{d}{dt}{i}_L\left(t\right)=\left[-0.242e^{-0.242t}\left(A\cos \left(0.159t\right)+B\sin \left(0.159t\right)\right)+0.159e^{-0.242t}\left(-A\cos \left(0.159t\right)+B\sin \left(0.159t\right)\right)\right]\mathrm{.................}\left(3\right)\end{array}$

Substituting 0 for t,

  $\begin{array}{l}{\frac{d}{dt}{i}_L\left(t\right)|}_{t=0}=\left[-0.242\left(1\right)\left(A\left(1\right)+B\left(0\right)\right)+0.159\left(1\right)\left(-A\left(0\right)+B\left(1\right)\right)\right]\\ {\frac{d}{dt}{i}_L\left(t\right)|}_{t=0}=-0.242A+0.159B\mathrm{....................}\left(4\right)\end{array}$

Applying KVL in the loop with current $I_1$ ,

  $\begin{array}{l}2I_1\left(t\right)+v_c\left(t\right)+5\frac{d{I}_1}{dt}=0\\ \text{Substituting }{i}_L\left(t\right)\text{for }{I}_1\left(t\right),\\ 2i_L\left(t\right)+v_c\left(t\right)+5\frac{d}{dt}{i}_L\left(t\right)=\mathrm{0................}\left(5\right)\\ \text{Substituting }0\text{for }t,\\ 2i_L\left(0\right)+v_c\left(0\right)+5{\frac{d}{dt}{i}_L\left(t\right)|}_{t=0}=0\end{array}$

Substituting the values,

  $\begin{array}{l}2\left(2\right)+6+5{\frac{d}{dt}{i}_L\left(t\right)|}_{t=0}=0\\ 5{\frac{d}{dt}{i}_L\left(t\right)|}_{t=0}=-10\\ {\frac{d}{dt}{i}_L\left(t\right)|}_{t=0}=-2\end{array}$

Substituting this in equation 4,

  $\begin{array}{l}-2=-0.242A+0.159B\\ -2=-0.242\left(2\right)+0.159B\\ B=\frac{-2+0.484}{0.159}\\ B=-\frac{1.516}{0.159}\\ B=-9.53\end{array}$

Substituting A and B in equation 2,

  $i_L\left(t\right)=e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)\text{for 0}\le \text{t}\le 5\text{s}\mathrm{..............}\left(6\right)$

Substituting A and B in equation 3,

  $\begin{array}{l}\frac{d}{dt}{i}_L\left(t\right)=\left[-0.242e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)+0.159e^{-0.242t}\left(-2\sin \left(0.159t\right)-9.53\cos \left(0.159t\right)\right)\right]\\ \frac{d}{dt}{i}_L\left(t\right)=e^{-0.242t}\left[-1.99\cos \left(0.159t\right)+1.98\sin \left(0.159t\right)\right]\end{array}$

Put these values in equation 5,

  $\begin{array}{l}0=2\left(e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)\right)+v_C\left(t\right)+5\left(e^{-0.242t}\left(-1.99\cos \left(0.159t\right)+1.98\sin \left(0.159t\right)\right)\right)\\ v_C\left(t\right)=-2\left(e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)\right)-5\left(e^{-0.242t}\left(-1.99\cos \left(0.159t\right)+1.98\sin \left(0.159t\right)\right)\right)\\ v_C\left(t\right)=e^{-0.242t}\left(6\cos \left(0.159t\right)+9.18\sin \left(0.159t\right)\right)\text{ for 0}\le \text{t}\le 5\text{s}\mathrm{...............}\left(7\right)\end{array}$

Substitute $t=5s$ in equation 6,

  $\begin{array}{l}{i}_L\left(5\right)=e^{-0.242\left(5\right)}\left(2\cos \left(0.159\left(5\right)\right)-9.53\sin \left(0.159\left(5\right)\right)\right)\\ i_L\left(5\right)=e^{-1.21}\left(2\left(0.7\right)-9.53\sin \left(0.713\right)\right)\\ i_L\left(5\right)=0.3\left(-5.395\right)\\ i_L\left(5\right)=-1.62\text{A}\end{array}$

Substitute $t=5s$ in equation 7,

  $\begin{array}{l}{v}_C\left(t\right)=e^{-0.242t}\left(6\cos \left(0.159t\right)+9.18\sin \left(0.159t\right)\right)\\ v_C\left(5\right)=e^{-0.242\left(5\right)}\left(6\cos \left(0.159\left(5\right)\right)+9.18\sin \left(0.159\left(5\right)\right)\right)\\ v_C\left(5\right)=e^{-1.21}\left(6\left(0.7\right)+9.18\left(0.713\right)\right)\\ v_C\left(5\right)=0.3\left(10.76\right)\\ v_C\left(5\right)=3.23\text{V}\end{array}$

Reclose the circuit. Now, the initial current in the inductor is represented as a voltage source of $5i_L\left(5\right)\text{V}$ ,

Thus,

  $\begin{array}{l}5i_L\left(5\right)=\left(5\right)\left(-1.62\right)\\ 5i_L\left(5\right)=-8.1\text{V}\end{array}$

The initial voltage across the capacitor is represented as a voltage source of $v_C\left(5\right)\text{V}$ ,

  $v_C\left(5\right)=3.23\text{V}$

The circuit can be drawn as

  

It can be observed that the voltage across the $2\Omega$ resistor is 6V. Thus,

  $\begin{array}{l}-2I_1=6\\ 2I_1=-6\end{array}$

Apply KVL in the loop with current $I_1$ ,

  $\begin{array}{l}2I_1+3\left(I_1-I_2\right)+5\frac{d{I}_1}{dt}-\left(-8.1\right)=0\\ \text{Substituting}-\frac{6}{s}\text{for }2I_1,\\ -6+3\left(I_1-I_2\right)+5\frac{d{I}_1}{dt}+8.1=0\\ 3I_1+5\frac{d{I}_1}{dt}-3I_2+2.1=0\\ I_2=\frac{3I_1+5\frac{d{I}_1}{dt}+2.1}{3}\end{array}$

Apply KVL in the loop with current $I_2$ ,

  $\begin{array}{l}3\left(I_2-I_1\right)+\frac{1}{4}{\displaystyle \int I_{2}dt}+3.23=0\\ \text{Differentiate with respect to t,}\\ \frac{d}{dt}\left(3\left(I_2-I_1\right)+\frac{1}{4}{\displaystyle \int I_{2}dt}+3.23\right)=0\\ 3\frac{d{I}_2}{dt}-3\frac{d{I}_1}{dt}+\frac{1}{4}{I}_2=0\\ \text{Substituting }{I}_2=\frac{3I_1+5\frac{d{I}_1}{dt}+2.1}{3},\\ 3\frac{d}{dt}\left(\frac{3I_1+5\frac{d{I}_1}{dt}+2.1}{3}\right)-3\frac{d{I}_1}{dt}+\frac{1}{4}\left(\frac{3I_1+5\frac{d{I}_1}{dt}+2.1}{3}\right)=0\\ 3\frac{d{I}_1}{dt}+5\frac{d^2{I}_1}{d{t}^2}-3\frac{d{I}_1}{dt}+\frac{1}{4}{I}_1+\frac{5}{12}\frac{d{I}_1}{dt}-\frac{2.1}{12}=0\\ 60\frac{d^2{I}_1}{d{t}^2}+5\frac{d{I}_1}{dt}+3I_1=2.1\\ \frac{d^2{I}_1}{d{t}^2}+\frac{1}{12}\frac{d{I}_1}{dt}+\frac{1}{20}{I}_1=\frac{2.1}{60}\end{array}$

Assuming $\frac{d}{dt}=D$,

  $\begin{array}{l}{D}^2{I}_1+\frac{1}{12}D{I}_1+\frac{1}{20}{I}_1=0\\ \left(D^2+\frac{1}{12}D+\frac{1}{20}\right)I_1=0\end{array}$

The auxiliary equation can be written as

  $D^2+\frac{1}{12}D+\frac{1}{20}=0$

The roots can be calculated as

  $\begin{array}{l}D=\frac{-\frac{1}{12}\pm \sqrt{{\left(\frac{1}{12}\right)}^2-\left(4\right)\left(1\right)\left(\frac{1}{20}\right)}}{2\left(1\right)}\\ D=-0.042+j0.220\end{array}$

Therefore, the solution of the differential equation will be in the form

  $I_1\left(t\right)=2.1+e^{-0.042\left(t-5\right)}\left(A\cos \left(0.220\left(t-5\right)\right)+B\sin \left(0.220\left(t-5\right)\right)\right)$

Current through the inductor is equal to the loop current,

  $i_L\left(t\right)=2.1+e^{-0.042\left(t-5\right)}\left(A\cos \left(0.220\left(t-5\right)\right)+B\sin \left(0.220\left(t-5\right)\right)\right)\mathrm{....................}\left(8\right)$

Using the initial conditions,

At t = 5,

  $\begin{array}{l}{i}_{L1}\left(5\right)=i_L\left(5\right)\\ i_{L1}\left(5\right)=-1.62\text{A}\end{array}$

Substituting 5 for t,

  $\begin{array}{l}{i}_{L1}\left(5\right)=2.1+A\\ -1.62=2.1+A\\ A=-1.62-2.1\\ A=-3.72\end{array}$

Differentiating equation 8 with respect to t,

  $\begin{array}{l}\frac{d}{dt}{i}_{L1}\left(t\right)=\frac{d}{dt}\left[e^{-0.042\left(t-5\right)}\left(A\cos \left(0.220t\right)+B\sin \left(0.220t\right)\right)\right]\\ \frac{d}{dt}{i}_{L1}\left(t\right)=\left[\begin{array}{l}-0.042e^{-0.042\left(t-5\right)}\left(A\cos \left(0.220\left(t-5\right)\right)+B\sin \left(0.220\left(t-5\right)\right)\right)+0.220e^{-0.042\left(t-5\right)}\\ \left(-A\sin 0.220\left(t-5\right)+B\cos 0.220\left(t-5\right)\right)\end{array}\right]\mathrm{.................}\left(9\right)\end{array}$

Substituting 5 for t,

  $\begin{array}{l}{\frac{d}{dt}{i}_{L1}\left(t\right)|}_{t=5}=\left[-0.042\left(0\right)\left(A\left(1\right)+B\left(0\right)\right)+0.220\left(1\right)\left(-A\left(0\right)+B\left(1\right)\right)\right]\\ {\frac{d}{dt}{i}_{L1}\left(t\right)|}_{t=5}=-0.042A+0.220B\mathrm{....................}\left(10\right)\end{array}$

The voltage across the inductor at t = 5s will be

  $\begin{array}{l}{v}_L\left(5\right)=5{\frac{d}{dt}{i}_{L1}\left(t\right)|}_{t=5}\\ 6-v_c\left(5\right)=5{\frac{d}{dt}{i}_{L1}\left(t\right)|}_{t=5}\\ \text{Substituting}{v}_c\left(5\right)=3.23\text{V}\\ 6-3.23\text{V}=5{\frac{d}{dt}{i}_{L1}\left(t\right)|}_{t=5}\\ 5{\frac{d}{dt}{i}_{L1}\left(t\right)|}_{t=5}=2.77\\ {\frac{d}{dt}{i}_{L1}\left(t\right)|}_{t=5}=\frac{2.77}{5}\\ {\frac{d{I}_1}{dt}|}_{t=5}=0.554\end{array}$

Substituting this in equation 10,

  $\begin{array}{l}0.554=-0.042A+0.220B\\ 0.554=-442\left(-3.72\right)+0.220B\\ 0.220B=0.554-0.15624\\ B=\frac{0.398}{0.220}\\ B=1.8\end{array}$

Substituting A and B in equation 8,

  $i_{L1}\left(t\right)=2.1+e^{-0.042\left(t-5\right)}\left(-3.72\cos \left(0.220\left(t-5\right)\right)+1.8\sin \left(0.220\left(t-5\right)\right)\right)\text{ for t }\ge \text{5s}$

The complete solution can be calculated as

  $\begin{array}{l}{i}_{Lo}\left(t\right)=i_L\left(t\right)+i_{L1}\left(t\right)\\ \text{Substituting the values,}\\ i_{Lo}\left(t\right)=\left[\begin{array}{l}{e}^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)\\ +2.1+e^{-0.042\left(t-5\right)}\left(-3.72\cos \left(0.220\left(t-5\right)\right)+1.8\sin \left(0.220\left(t-5\right)\right)\right)\end{array}\right]\\ i_{Lo}\left(t\right)=\left[\begin{array}{l}2.1+e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)+\\ e^{-0.042\left(t-5\right)}\left(-3.72\cos \left(0.220\left(t-5\right)\right)+1.8\sin \left(0.220\left(t-5\right)\right)\right)\text{A}\end{array}\right]\end{array}$

The 6V is always connected across the 2ohm resistor whenever the switch is closed. Therefore, the voltage across the 2ohm resistor will be

  $V_1=6\text{V}$

When the switch is opened, the current through the inductor flows through the 2ohm resistor. Thus, the voltage across the 2ohm resistor, when the switch is open will be

  $V_2=2i_L\left(t\right)$

Substituting the value of $i_L\left(t\right)$ ,

  $\begin{array}{l}{V}_2=2\left[e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)\right]\\ V_2=2e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)\end{array}$

The voltage across the 2ohm resistor will be

  $\begin{array}{l}{V}_{2\Omega }=V_1+V_2\\ V_{2\Omega }=6+2e^{-0.242t}\left(2\cos \left(0.159t\right)-9.53\sin \left(0.159t\right)\right)\end{array}$