Chapter 7, Problem 80P

In the circuit of Fig. 7.144, find the value of i_o for all values of 0 < t.

To determine

Find the current $i_o$ through the $10\Omega$ resistor in the given circuit of Figure 7.144.

Answer to Problem 80P

The current $i_o$ through the $10\Omega$ resistor is $\left[500+625e^{-2t}\right]u\left(t\right)\text{mA}$.

Explanation of Solution

Given data:

Refer to Figure 7.144 in the textbook.

The value of capacitance $\left(C\right)$ is $50\text{mF}$.

The source voltage $v_s$ is $25\left[1-u\left(t\right)\right]\text{V}$.

The current source $i_s$ is 1 A.

Formula used:

Write the general expression to find the complete voltage response for an RC circuit.

$v_C\left(t\right)=v_C\left(\infty \right)+\left[v_C\left(0\right)-v_C\left(\infty \right)\right]e^{-\frac{t}{\tau }}$ (1)

Here,

$\tau$ is the time constant for the RC circuit,

$v_C\left(0\right)$ is the initial capacitor voltage, and

$v_C\left(\infty \right)$ is the final capacitor voltage.

Write the expression to find the time constant for an RC circuit.

$\tau =R_{\text{Th}}C$ (2)

Here,

$R_{\text{Th}}$ is the Thevenin resistance, and

C is the capacitance of the capacitor.

Write the general expression for the unit step function.

$u(t)=\{\begin{array}{l}0,t<0\\ 1,t>0\end{array}$ (3)

Calculation:

The given source voltage is,

$v_s=25\left[1-u\left(t\right)\right]\text{V}$ (4)

Apply the unit step function in equation (3) to equation (4).

$v_s=\{\begin{array}{l}25\text{V},t<0\\ 0\text{V},t>0\end{array}$

For $t<0$:

The given Figure 7.144 is redrawn as shown in Figure 1.

In Figure 1, the capacitor reaches steady state and it will acts as an open circuit. The initial voltage across the capacitor is denoted by $v_C\left(0\right)$.

Apply Kirchhoff’s current law at node $v_o$ in Figure 1.

$\begin{array}{l}1=\frac{v_o-25}{10}+\frac{v_o}{10}\\ 1=\frac{v_o}{10}-\frac{25}{10}+\frac{v_o}{10}\\ 1=\frac{2v_o}{10}-\frac{25}{10}\\ 1=\frac{1}{10}\left(2v_o-25\right)\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}10=2v_o-25\\ 2v_o=10+25\\ v_o=\frac{10+25}{2}\\ v_o=17.5\text{V}\end{array}$

In Figure 1, the initial voltage across the capacitor $v_C\left(0\right)$ is equal to the nodal voltage $v_o$. Therefore, $v_C\left(0\right)=17.5\text{V}$.

For $t>0$:

In Figure 2, the voltage source is equal to zero (or a short circuit). Now, the final voltage across the capacitor is represented by $v_C\left(\infty \right)$.

Apply Kirchhoff’s current law at node $v_o$ in Figure 2.

$\begin{array}{l}1=\frac{v_o}{10}+\frac{v_o}{10}\\ 1=\frac{2v_o}{10}\end{array}$

Rearrange the equation as follows,

$\begin{array}{l}10=2v_o\\ v_o=\frac{10}{2}\\ v_o=5\text{V}\end{array}$

In Figure 2, the final voltage across the capacitor $v_C\left(\infty \right)$ is equal to the nodal voltage $v_o$. Therefore, $v_C\left(\infty \right)=5\text{V}$.

Figure 3 shows the Thevenin resistance at the capacitor terminal.

In Figure 3, the Thevenin resistance is calculated as follows.

$\begin{array}{c}{R}_{\text{Th}}=\left(10\Omega \left|\right|10\Omega \right)+5\Omega \\ =\frac{10\Omega \times 10\Omega }{10\Omega +10\Omega }+5\Omega \\ =5\Omega +5\Omega \\ =10\Omega \end{array}$

Substitute $10\Omega$ for $R_{\text{Th}}$ and $50\text{mF}$ for C in equation (2) to find the time constant $\tau$.

$\tau =\left(10\Omega \right)\left(50\text{mF}\right)$ (5)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{A}\cdot \text{s}}{\text{V}}$ for F in equation (5) to find the time constant $\tau$ in seconds.

$\begin{array}{c}\tau =\left(10\frac{\text{V}}{\text{A}}\right)\left(50\text{m}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\\ =500\text{ms}\\ =500\times {10}^{-3}\text{s}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =0.5\text{s}\end{array}$

Substitute $17.5\text{V}$ for $v_C\left(0\right)$, $5\text{V}$ for $v_C\left(\infty \right)$, and $0.5\text{s}$ for $\tau$ in equation (1) to find the output voltage across the capacitor $v_C\left(t\right)$ in volts.

$\begin{array}{c}{v}_C\left(t\right)=5\text{V}+\left[17.5\text{V}-5\text{V}\right]e^{-\frac{t}{0.5\text{s}}}\\ =5\text{V}+\left[12.5\text{V}\right]e^{-\frac{t}{0.5\text{s}}}\\ =5+12.5e^{-2t}\text{V}\end{array}$

Figure 4 shows the modified circuit diagram.

Apply Kirchhoff’s current law at node $v_o$ in Figure 4.

$\begin{array}{l}1=\frac{v_o}{10}+\frac{v_o}{10}+\frac{v_o-v_C\left(t\right)}{5}\\ 1=\frac{v_o}{10}+\frac{v_o}{10}+\frac{v_o}{5}-\frac{v_C\left(t\right)}{5}\\ 1=v_o\left(\frac{1}{10}+\frac{1}{10}+\frac{1}{5}\right)-\frac{v_C\left(t\right)}{5}\\ 1=0.4v_o-0.2v_C\left(t\right)\end{array}$

Substitute $5+12.5e^{-2t}\text{V}$ for $v_C\left(t\right)$ to find the voltage $v_o$ in volts.

$\begin{array}{l}1=0.4v_o-0.2\left(5+12.5e^{-2t}\right)\\ 1=0.4v_o-1-2.5e^{-2t}\\ 0.4v_o=1+1+2.5e^{-2t}\\ 0.4v_o=2+2.5e^{-2t}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{v}_o=\frac{2+2.5e^{-2t}}{0.4}\\ =5+6.25e^{-2t}\text{V}\end{array}$

Therefore, the current $i_o$ through the $10\Omega$ resistor is calculated by using Ohm’s law.

$i_o=\frac{v_o}{10\Omega }$

Substitute $5+6.25e^{-2t}\text{V}$ for $v_o$ to find the current $i_o$ in amperes.

$\begin{array}{c}{i}_o=\frac{5+6.25e^{-2t}\text{V}}{10\Omega }\\ =0.5+0.625e^{-2t}\frac{\text{V}}{\Omega }\\ =0.5+0.625e^{-2t}\text{A}\left\{\because 1\text{A}=\frac{\text{1}\text{V}}{1\Omega }\right\}\end{array}$

Convert the unit A to mA.

$\begin{array}{c}{i}_o=\left[0.5+0.625e^{-2t}\text{A}\right]\times {10}^3\times {10}^{-3}\\ =\left[0.5+0.625e^{-2t}\text{mA}\right]\times {10}^3\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

$i_o=\left[500+625e^{-2t}\right]\text{mA}$ (6)

Apply the unit step function in equation (3) to equation (6).

$\begin{array}{c}{i}_o=\left(\left[500+625e^{-2t}\right]\text{mA}\right)u\left(t\right)\\ =\left[500+625e^{-2t}\right]u\left(t\right)\text{mA}\end{array}$

PSpice Simulation:

For $t<0$:

Draw the circuit diagram in PSpice as shown in Figure 5.

Save the circuit and provide the Simulation Settings as shown in Figure 6.

Now run the simulation and the results will be displayed as shown in Figure 7 by enabling “Enable Bias Voltage Display” icon.

From Figure 7, the initial voltage across the capacitor is 17.5 V.

For $t>0$:

Draw the circuit diagram in PSpice as shown in Figure 8.

Now run the simulation and the results will be displayed as shown in Figure 8 by enabling “Enable Bias Voltage Display” icon.

From Figure 9, the final voltage across the capacitor is 5 V.

Draw the circuit diagram in PSpice as shown in Figure 10.

Now run the simulation and the results will be displayed as shown in Figure 11 by enabling “Enable Bias Voltage Display” icon and place the “Current Marker”

The SCHEMATIC1 dialog box is also opened with simulation result as shown in Figure 12.

Therefore, the plot of current through the $10\Omega$ resistor is obtained.

Conclusion:

Thus, the current $i_o$ through the $10\Omega$ resistor is $\left[500+625e^{-2t}\right]u\left(t\right)\text{mA}$.