Chapter 7, Problem 50P

In the circuit of Fig. 7.117, find i_x for t > 0. Let R₁ = R₂ = 1 kΩ, R₃ = 2 kΩ, and C = 0.25 mF.

Figure 7.117

To determine

Calculate the current $i_x$ through the resistance $R_1$ for $t>0$ in the given circuit of Figure 7.117.

Answer to Problem 50P

The current $i_x$ through the resistance $R_1$ is $7.5\left(3-e^{-4t}\right)\text{mA}$ for $t>0$.

Explanation of Solution

Given data:

Refer to Figure 7.117 in the textbook.

The source current $\left(i_s\right)$ is $30\text{mA}$

The value of resistances are $R_1=R_2=1\text{k}\Omega$ and $R_3=2\text{k}\Omega$.

The value of capacitance $\left(C\right)$ is $0.25\text{mF}$.

Formula used:

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

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

Here,

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

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

$v\left(\infty \right)$ 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.

Calculation:

The given Figure 7.117 is redrawn as shown in Figure 1.

For $t<0$:

The switch is kept open at this condition. Therefore, the initial capacitor voltage $v\left(0\right)$ is equal to zero. That is,

$v\left(0\right)=0\text{V}$

For $t>0$:

In Figure 1, the current source $30\text{mA}$ with parallel resistance $R_1=1\text{k}\Omega$ is transformed into voltage source with series resistance $R_1=1\text{k}\Omega$ using source transformation.

That is,

$\begin{array}{c}V=\left(30\text{mA}\right)\left(1\text{k}\Omega \right)\\ =\left(30\times {10}^{-3}\text{A}\right)\left(1\times {10}^3\Omega \right)\left\{\because 1\text{m}={10}^{-3},1\text{k}={10}^3\right\}\\ =30\text{V}\left\{\because 1\text{V}=1\text{A}\cdot 1\Omega \right\}\end{array}$

Figure 2 shows the modified circuit diagram when $t>0$.

In Figure 2, the final capacitor voltage $v\left(\infty \right)$ is calculated by using voltage division rule.

$\begin{array}{c}v\left(\infty \right)=\left(30\text{V}\right)\left(\frac{2\text{k}\Omega }{1\text{k}\Omega +1\text{k}\Omega +2\text{k}\Omega }\right)\\ =\left(30\text{V}\right)\left(\frac{2}{4}\right)\\ =15\text{V}\end{array}$

Figure 2 shows the Thevenin resistance $R_{\text{Th}}$ at the capacitor terminal.

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

$\begin{array}{c}{R}_{\text{Th}}=\left(1\text{k}\Omega +1\text{k}\Omega \right)\left|\right|2\text{k}\Omega \\ =2\text{k}\Omega \left|\right|2\text{k}\Omega \\ =\frac{2\text{k}\Omega \times 2\text{k}\Omega }{2\text{k}\Omega +2\text{k}\Omega }\\ =1\text{k}\Omega \end{array}$

Substitute $1\text{k}\Omega$ for $R_{\text{Th}}$ and $0.25\text{mF}$ for C in equation (2) to find the time constant $\tau$.

$\tau =\left(1\text{k}\Omega \right)\left(0.25\text{mF}\right)$

$\tau =\left(1\times {10}^3\Omega \right)\left(0.25\times {10}^{-3}\text{F}\right)\left\{\because 1\text{m}={10}^{-3},1\text{k}={10}^3\right\}$ (3)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{A}\cdot \text{s}}{\text{V}}$ for F in equation (3) to find the time constant $\tau$ in seconds.

$\begin{array}{c}\tau =\left(1\times {10}^3\frac{\text{V}}{\text{A}}\right)\left(0.25\times {10}^{-3}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\\ =0.25\text{s}\end{array}$

Substitute $0\text{V}$ for $v\left(0\right)$, $15\text{V}$ for $v\left(\infty \right)$, and $0.25\text{s}$ for $\tau$ in equation (1) to find the output voltage across the capacitor $v\left(t\right)$ in volts.

$\begin{array}{c}v\left(t\right)=15\text{V}+\left[0\text{V}-15\text{V}\right]e^{-\frac{t}{0.25\text{s}}}\\ =15\text{V}-\left[15\text{V}\right]e^{-4t}\end{array}$

$v\left(t\right)=15\left(1-e^{-4t}\right)\text{V}$ (4)

Figure 4 shows the modified circuit diagram to find the current $i_x$.

In Figure 4, apply Kirchhoff’s current law at node 1.

$i_x=30\text{mA}-i_T$ (5)

In Figure 4, apply Kirchhoff’s current law at node $v$.

$i_T=\frac{v}{R_3}+C\frac{dv}{dt}$ (6)

Substitute $15\left(1-e^{-4t}\right)\text{V}$ for v, $2\text{k}\Omega$ for $R_3$, $0.25\text{mF}$ for C in equation (6) to find the current $i_T$ in amperes.

$\begin{array}{c}{i}_T=\frac{15\left(1-e^{-4t}\right)\text{V}}{2\text{k}\Omega }+\left(0.25\text{mF}\right)\frac{d}{dt}\left(15\left(1-e^{-4t}\right)\text{V}\right)\\ =\frac{15\left(1-e^{-4t}\right)\text{V}}{2\times {10}^3\Omega }+\left(0.25\times {10}^{-3}\text{F}\right)\frac{d}{dt}\left(15\left(1-e^{-4t}\right)\text{V}\right)\left\{\because 1\text{m}={10}^{-3},1\text{k}={10}^3\right\}\\ =\frac{15\left(1-e^{-4t}\right)\text{V}}{2\times {10}^3\Omega }+\left(0.25\times {10}^{-3}\text{F}\right)\left(15\left(0-\left(-4\right)e^{-4t}\right)\frac{\text{V}}{\text{s}}\right)\\ =7.5\times {10}^{-3}\left(1-e^{-4t}\right)\text{A}+\left(0.25\times {10}^{-3}\text{F}\right)\left(60e^{-4t}\frac{\text{V}}{\text{s}}\right)\left\{\because 1\text{A}=\frac{1\text{V}}{1\Omega }\right\}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{i}_T=7.5\times {10}^{-3}\left(1-e^{-4t}\right)\text{A}+\left(0.25\times {10}^{-3}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\left(60e^{-4t}\frac{\text{V}}{\text{s}}\right)\left\{\because 1\text{F}=\frac{\text{1A}\cdot 1\text{s}}{\text{V}}\right\}\\ =7.5\times {10}^{-3}\text{A}-7.5\times {10}^{-3}{e}^{-4t}\text{A}+\text{15}\times {10}^{-3}{e}^{-4t}\text{A}\\ =7.5\times {10}^{-3}\text{A}+7.5\times {10}^{-3}{e}^{-4t}\text{A}\end{array}$

Substitute $7.5\times {10}^{-3}\text{A+}7.5\times {10}^{-3}{e}^{-4t}\text{A}$ for $i_T$ in equation (5) to find the current $i_x$ in amperes.

$\begin{array}{c}{i}_x=30\text{mA}-\left(7.5\times {10}^{-3}\text{A}+7.5\times {10}^{-3}{e}^{-4t}\text{A}\right)\\ =30\text{mA}-7.5\times {10}^{-3}\text{A}-7.5\times {10}^{-3}{e}^{-4t}\text{A}\\ =30\text{mA}-7.5\text{mA}-7.5e^{-4t}\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =22.5\text{mA}-7.5e^{-4t}\text{mA}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}{i}_x=\left(22.5-7.5e^{-4t}\right)\text{mA}\\ =7.5\left(3-e^{-4t}\right)\text{mA}\end{array}$

Conclusion:

Thus, the current $i_x$ through the resistance $R_1$ is $7.5\left(3-e^{-4t}\right)\text{mA}$ for $t>0$.