Chapter 6, Problem 31P

If v(0) = 0, find v(t), i₁(t), and i₂(t) in the circuit of Fig. 6.63.

Figure 6.63

For Prob. 6.31.

To determine

Find the expression for voltage $v\left(t\right)$, current $i_1\left(t\right)$ and $i_2\left(t\right)$ in the circuit of Figure 6.63.

Answer to Problem 31P

The expression of voltage $v\left(t\right)$, current $i_1\left(t\right)$ and $i_2\left(t\right)$ is,

$v\left(t\right)=\{\begin{array}{l}1.5t^2\text{kV},0\text{s}<t<1\text{s}\\ \left(3t-1.5\right)\text{kV},1\text{s}<t<3\text{s}\\ \left(0.75t^2-7.5t+23.25\right)\text{kV,}3\text{s}<t<5\text{s}\end{array}$

$i_1\left(t\right)=\{\begin{array}{l}18t\text{mA},0\text{s}<t<1\text{s}\\ 18\text{mA},1\text{s}<t<3\text{s}\\ \left(9t-45\right)\text{mA,}3\text{s}<t<5\text{s}\end{array}$

$i_2\left(t\right)=\{\begin{array}{l}12t\text{mA},0\text{s}<t<1\text{s}\\ 12\text{mA},1\text{s}<t<3\text{s}\\ \left(6t-30\right)\text{mA,}3\text{s}<t<5\text{s}\end{array}$.

Explanation of Solution

Given data:

Refer to Figure 6.63 in the textbook.

The initial voltage at time $t_0$ $\left(v\left(0\right)\right)$ is $0\text{V}$.

Formula used:

Write the expression to calculate the straight line equation for two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$.

$\left(y-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ (1)

Refer to Figure 6.63 in the textbook.

From the given graph, substitute $t$ for $x$ and $i_s$ for $y$ in equation (1).

$\left(i_s-y_1\right)=\frac{y_2-y_1}{x_2-x_1}\left(t-x_1\right)$ (2)

Write the expression to calculate the voltage across the capacitor.

$v\left(t\right)=\frac{1}{C}{\displaystyle \underset{0}{\overset{t}{\int }}{i}_s\left(t\right)dt+v\left(t_0\right)}$ (3)

Here,

$C_2$ is the value of the capacitor 2,

$i_s\left(t\right)$ is the current through the capacitors and

$v\left(t_0\right)$ is the voltage across the capacitor $C$ at time $t_0$.

Calculation:

The given circuit is redrawn as Figure 1.

Refer to Figure 1, the capacitors $C_1$ and $C_2$ are connected in parallel form.

Write the expression to calculate the equivalent capacitance for the parallel connected capacitors $C_1$ and $C_2$.

$C=C_1+C_2$ (4)

Here,

$C_1$ is the value of the capacitor 1, and

$C_2$ is the value of the capacitor 2.

Substitute $6\mu \text{F}$ for $C_1$ and $4\mu \text{F}$ for $C_2$ in equation (4) to find $C$.

$\begin{array}{c}C=6\mu \text{F}+4\mu \text{F}\\ =10\mu \text{F}\end{array}$

The given current waveform is redrawn as Figure 2.

Refer to Figure 2, split up the time period as three divisions $0\text{s}<t<1\text{s}$, $1\text{s}<t<3\text{s}$ and $3\text{s}<t<5\text{s}$ to find the respective current value.

Case (i): $0\text{s}<t<1\text{s}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(0\text{s},\text{0}\text{mA}\right)$ and $\left(1\text{s},3\text{0}\text{mA}\right)$.

Substitute $0\text{s}$ for $x_1$, $\text{0}\text{mA}$ for $y_1$, $1\text{s}$ for $x_2$ and $\text{30}\text{mA}$ for $y_2$ in equation (2) to find $i_s\left(t\right)$.

$\left(i_s\left(t\right)-\text{0}\text{mA}\right)=\frac{\text{30}\text{mA}-\text{0}\text{mA}}{1\text{s}-0\text{s}}\left(t-0\text{s}\right)$

Simplify the equation to find $i_s\left(t\right)$.

$\begin{array}{c}{i}_s\left(t\right)=\frac{\text{30}\text{mA}}{1\text{s}}\left(t\right)\\ =30t\text{mA}\end{array}$

Case (ii): $1\text{s}<t<3\text{s}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(1\text{s},3\text{0}\text{mA}\right)$ and $\left(3\text{s},30\text{mA}\right)$.

Substitute $1\text{s}$ for $x_1$, $\text{30}\text{mA}$ for $y_1$, $3\text{s}$ for $x_2$ and $\text{30}\text{mA}$ for $y_2$ in equation (2) to find $i_s\left(t\right)$.

$\begin{array}{c}\left(i_s\left(t\right)-\text{30}\text{mA}\right)=\frac{\text{30}\text{mA}-\text{30}\text{mA}}{3\text{s}-1\text{s}}\left(t-1\text{s}\right)\\ =\frac{\text{0}\text{mA}}{2\text{s}}\left(t-1\text{s}\right)\\ =0\end{array}$

Simplify the equation to find $i_s\left(t\right)$.

$i_s\left(t\right)=30\text{mA}$

Case (iii): $3\text{s}<t<5\text{s}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(3\text{s},-30\text{mA}\right)$ and $\left(5\text{s},0\text{mA}\right)$.

Substitute $3\text{s}$ for $x_1$, $-\text{30}\text{mA}$ for $y_1$, $5\text{s}$ for $x_2$ and $\text{0}\text{mA}$ for $y_2$ in equation (2) to find $i_s\left(t\right)$.

$\begin{array}{c}\left(i_s\left(t\right)+\text{30}\text{mA}\right)=\frac{\text{0}\text{mA}+\text{30}\text{mA}}{5\text{s}-3\text{s}}\left(t-3\text{s}\right)\\ =\frac{\text{30}\text{mA}}{2\text{s}}\left(t-3\text{s}\right)\\ =\text{15}\text{mA}\left(t-3\text{s}\right)\\ =\text{15}t\text{mA}-\text{45}\text{mA}\end{array}$

Simplify the equation to find $i_s\left(t\right)$.

$\begin{array}{c}{i}_s\left(t\right)=\text{15}t\text{mA}-\text{45}\text{mA}-\text{30}\text{mA}\\ =15t\text{mA}-\text{75}\text{mA}\\ =\left(15t-75\right)\text{mA}\end{array}$

Therefore, the current function of the signal in Figure 2 is,

$i_s\left(t\right)=\{\begin{array}{l}30t\text{mA},0\text{s}<t<1\text{s}\\ 30\text{mA},1\text{s}<t<3\text{s}\\ \left(15t-75\right)\text{mA,}3\text{s}<t<5\text{s}\end{array}$

For $0\text{s}<t<1\text{s}$:

Substitute $10\mu \text{F}$ for $C$, $30t\text{mA}$ for $i_s\left(t\right)$ and $0\text{V}$ for $v\left(t_0\right)$ in equation (3) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\frac{1}{\left(10\mu \text{F}\right)}{\displaystyle \underset{0}{\overset{t}{\int }}\left(30t\text{mA}\right)dt+0\text{V}}\\ =\frac{30\times {\text{10}}^{-3}}{\left(10\times {10}^{-6}\right)}{\displaystyle \underset{0}{\overset{t}{\int }}\left(tdt\right)\frac{\text{A}}{\text{F}}\left\{\because 1\mu ={10}^{-6},1\text{m}={\text{10}}^{-3}\right\}}\\ =\left(3\times {10}^3\right){\left[\frac{t^2}{2}\right]}_0^t\frac{\text{A}\cdot \text{s}}{\text{F}}\\ =\left(1.5\times {10}^3\right)\left[t^2-0^2\right]\text{V}\left\{\because 1\text{V}=\frac{\text{1A}\cdot 1\text{s}}{\text{1F}}\right\}\end{array}$

Simplify the equation to find $v\left(t\right)$.

$v\left(t\right)=\left(1.5\times {10}^3\right)t^2\text{V}$

$v\left(t\right)=1.5t^2\text{kV}\left\{\because 1\text{k}={\text{10}}^3\right\}$ (5)

Substitute $1$ for $t$ in equation (5) to find $v\left(1\right)$.

$\begin{array}{c}v\left(1\right)=1.5{\left(1\right)}^2\text{kV}\\ =1.5\text{kV}\end{array}$

For $1\text{s}<t<3\text{s}$:

Write the expression to calculate the capacitor voltage depends on the past history of the capacitor voltage.

$v\left(t\right)=\frac{1}{C}{\displaystyle \underset{0}{\overset{t}{\int }}{i}_s\left(t\right)dt+v\left(1\right)}$ (6)

Here,

$v\left(1\right)$ is the voltage across the capacitor $C$ at time $1\text{s}$.

Substitute $10\mu \text{F}$ for $C$, $30\text{mA}$ for $i_s\left(t\right)$ and $1.5\text{kV}$ for $v\left(1\right)$ in equation (6) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\frac{1}{\left(10\mu \text{F}\right)}{\displaystyle \underset{1}{\overset{t}{\int }}\left(30\text{mA}\right)dt+1.5\text{kV}}\\ =\frac{1}{\left(10\times {10}^{-6}\right)}\left(30\right){\displaystyle \underset{1}{\overset{t}{\int }}\left(dt\right){\text{10}}^{-3}\frac{\text{A}}{\text{F}}+1.5\text{kV}\left\{\because 1\mu ={10}^{-6},1\text{m}={\text{10}}^{-3}\right\}}\\ =\left(3\times {10}^3\right){\left[t\right]}_1^t\frac{\text{A}\cdot \text{s}}{\text{F}}+1.5\text{kV}\\ =\left(3\times {10}^3\right)\left[t-1\right]\text{V}+1.5\text{kV}\left\{\because 1\text{V}=\frac{\text{1A}\cdot 1\text{s}}{\text{1F}}\right\}\end{array}$

Simplify the equation to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\left(3t-3\right)\text{kV}+1.5\text{kV}\left\{\because 1\text{k}={\text{10}}^3\right\}\\ =3t\text{kV}-3\text{kV}+1.5\text{kV}\\ =3t\text{kV}-1.5\text{kV}\end{array}$

$v\left(t\right)=\left(3t-1.5\right)\text{kV}$ (7)

Substitute $3$ for $t$ in equation (7) to find $v\left(3\right)$.

$\begin{array}{c}v\left(3\right)=\left(3\left(3\right)-1.5\right)\text{kV}\\ =\left(9-1.5\right)\text{kV}\\ =7.5\text{kV}\end{array}$

For $3\text{s}<t<5\text{s}$:

Write the expression to calculate the capacitor voltage depends on the past history of the capacitor voltage.

$v\left(t\right)=\frac{1}{C}{\displaystyle \underset{0}{\overset{t}{\int }}{i}_s\left(t\right)dt+v\left(3\right)}$ (8)

Here,

$v\left(3\right)$ is the voltage across the capacitor $C$ at time $3\text{s}$.

Substitute $10\mu \text{F}$ for $C$, $\left(15t-75\right)\text{mA}$ for $i_s\left(t\right)$ and $7.5\text{kV}$ for $v\left(3\right)$ in equation (8) to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\frac{1}{\left(10\mu \text{F}\right)}{\displaystyle \underset{3}{\overset{t}{\int }}\left(\left(15t-75\right)\text{mA}\right)dt+7.5\text{kV}}\\ =\frac{1}{\left(10\times {10}^{-6}\right)}{\displaystyle \underset{3}{\overset{t}{\int }}\left(\left(15t-75\right)dt\right){\text{10}}^{-3}\frac{\text{A}}{\text{F}}+7.5\text{kV}\left\{\because 1\mu ={10}^{-6},1\text{m}={\text{10}}^{-3}\right\}}\\ =\left(100\right){\left[\frac{15t^2}{2}-75t\right]}_3^t\frac{\text{A}\cdot \text{s}}{\text{F}}+7.5\text{kV}\\ =100\left[\frac{15t^2}{2}-75t-\frac{15{\left(3\right)}^2}{2}+75\left(3\right)\right]\text{V}+7.5\text{kV}\left\{\because 1\text{V}=\frac{\text{1A}\cdot 1\text{s}}{\text{1F}}\right\}\end{array}$

Simplify the equation to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=100\left[7.5t^2-75t-67.5+225\right]\text{V}+7.5\text{kV}\\ =100\left[7.5t^2-75t+157.5\right]\text{V}+7.5\text{kV}\\ =1000\left[0.75t^2-7.5t+15.75\right]\text{V}+7.5\text{kV}\\ =\left[0.75t^2-7.5t+15.75\right]\times {10}^3\text{V}+7.5\text{kV}\end{array}$

Simplify the equation to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=\left[0.75t^2-7.5t+15.75\right]\text{kV}+7.5\text{kV}\left\{\because 1\text{k}={10}^3\right\}\\ =0.75t^2\text{kV}-7.5t\text{kV}+15.75\text{kV}+7.5\text{kV}\\ =0.75t^2\text{kV}-7.5t\text{kV}+23.25\text{kV}\\ =\left(0.75t^2-7.5t+23.25\right)\text{kV}\end{array}$

Therefore, the expression of the voltage $v\left(t\right)$ is,

$v\left(t\right)=\{\begin{array}{l}1.5t^2\text{kV},0\text{s}<t<1\text{s}\\ \left(3t-1.5\right)\text{kV},1\text{s}<t<3\text{s}\\ \left(0.75t^2-7.5t+23.25\right)\text{kV,}3\text{s}<t<5\text{s}\end{array}$

Write the expression to calculate the current through the capacitor $C_1$.

$i_1\left(t\right)=C_1\frac{dv\left(t\right)}{dt}$ (9)

For $0\text{s}<t<1\text{s}$:

Substitute $6\mu \text{F}$ for $C$ and $1.5t^2\text{kV}$ for $v\left(t\right)$ in equation (9) to find $i_1\left(t\right)$.

$\begin{array}{c}{i}_1\left(t\right)=\left(6\mu \text{F}\right)\frac{d\left(1.5t^2\text{kV}\right)}{dt}\\ =\left(6\times {10}^{-6}\right)\left(1.5\right)\left({10}^3\right)\frac{d\left(t^2\right)}{dt}\text{F}\cdot \text{V}\left\{\because 1\mu ={10}^{-6},1\text{k}={\text{10}}^3\right\}\\ =\left(9\times {10}^{-3}\right)\left(2t\right)\frac{\text{F}\cdot \text{V}}{\text{s}}\\ =\left(18t\times {10}^{-3}\right)\text{A}\left\{\because 1\text{A}=\frac{\text{1F}\cdot 1\text{V}}{\text{1s}}\right\}\end{array}$

Simplify the equation to find $i_1\left(t\right)$.

$i_1\left(t\right)=18t\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}$

For $1\text{s}<t<3\text{s}$:

Substitute $6\mu \text{F}$ for $C$ and $\left(3t-1.5\right)\text{kV}$ for $v\left(t\right)$ in equation (9) to find $i_1\left(t\right)$.

$\begin{array}{c}{i}_1\left(t\right)=\left(6\mu \text{F}\right)\frac{d\left(\left(3t-1.5\right)\text{kV}\right)}{dt}\\ =\left(6\times {10}^{-6}\right)\left({10}^3\right)\frac{d\left(3t-1.5\right)}{dt}\text{F}\cdot \text{V}\left\{\because 1\mu ={10}^{-6},1\text{k}={\text{10}}^3\right\}\\ =\left(6\times {10}^{-3}\right)\left(3-0\right)\frac{\text{F}\cdot \text{V}}{\text{s}}\\ =\left(6\times {10}^{-3}\right)\left(3\right)\text{A}\left\{\because 1\text{A}=\frac{\text{1F}\cdot 1\text{V}}{\text{1s}}\right\}\end{array}$

Simplify the equation to find $i_1\left(t\right)$.

$\begin{array}{c}{i}_1\left(t\right)=\left(18\times {10}^{-3}\right)\text{A}\\ =18\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

For $3\text{s}<t<5\text{s}$:

Substitute $6\mu \text{F}$ for $C$ and $\left(0.75t^2-7.5t+23.25\right)\text{kV}$ for $v\left(t\right)$ in equation (9) to find $i_1\left(t\right)$.

$\begin{array}{c}{i}_1\left(t\right)=\left(6\mu \text{F}\right)\frac{d\left(\left(0.75t^2-7.5t+23.25\right)\text{kV}\right)}{dt}\\ =\left(6\times {10}^{-6}\right)\left({10}^3\right)\frac{d\left(0.75t^2-7.5t+23.25\right)}{dt}\text{F}\cdot \text{V}\left\{\because 1\mu ={10}^{-6},1\text{k}={\text{10}}^3\right\}\\ =\left(6\times {10}^{-3}\right)\left(1.5t-7.5+0\right)\frac{\text{F}\cdot \text{V}}{\text{s}}\\ =\left(6\times {10}^{-3}\right)\left(1.5t-7.5\right)\text{A}\left\{\because 1\text{A}=\frac{\text{1F}\cdot 1\text{V}}{\text{1s}}\right\}\end{array}$

Simplify the equation to find $i_1\left(t\right)$.

$\begin{array}{c}{i}_1\left(t\right)=\left(6\right)\left(1.5t-7.5\right)\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\left(9t-45\right)\text{mA}\end{array}$

Therefore, the expression of current through the capacitor $C_1$ is,

$i_1\left(t\right)=\{\begin{array}{l}18t\text{mA},0\text{s}<t<1\text{s}\\ 18\text{mA},1\text{s}<t<3\text{s}\\ \left(9t-45\right)\text{mA,}3\text{s}<t<5\text{s}\end{array}$

Write the expression to calculate the current through the capacitor $C_2$.

$i_2\left(t\right)=C_2\frac{dv\left(t\right)}{dt}$ (10)

For $0\text{s}<t<1\text{s}$:

Substitute $4\mu \text{F}$ for $C$ and $1.5t^2\text{kV}$ for $v\left(t\right)$ in equation (10) to find $i_2\left(t\right)$.

$\begin{array}{c}{i}_2\left(t\right)=\left(4\mu \text{F}\right)\frac{d\left(1.5t^2\text{kV}\right)}{dt}\\ =\left(4\times {10}^{-6}\right)\left(1.5\right)\left({10}^3\right)\frac{d\left(t^2\right)}{dt}\text{F}\cdot \text{V}\left\{\because 1\mu ={10}^{-6},1\text{k}={\text{10}}^3\right\}\\ =\left(6\times {10}^{-3}\right)\left(2t\right)\frac{\text{F}\cdot \text{V}}{\text{s}}\\ =\left(12t\times {10}^{-3}\right)\text{A}\left\{\because 1\text{A}=\frac{\text{1F}\cdot 1\text{V}}{\text{1s}}\right\}\end{array}$

Simplify the equation to find $i_2\left(t\right)$.

$i_1\left(t\right)=12t\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}$

For $1\text{s}<t<3\text{s}$:

Substitute $4\mu \text{F}$ for $C$ and $\left(3t-1.5\right)\text{kV}$ for $v\left(t\right)$ in equation (10) to find $i_2\left(t\right)$.

$\begin{array}{c}{i}_2\left(t\right)=\left(4\mu \text{F}\right)\frac{d\left(\left(3t-1.5\right)\text{kV}\right)}{dt}\\ =\left(4\times {10}^{-6}\right)\left({10}^3\right)\frac{d\left(3t-1.5\right)}{dt}\text{F}\cdot \text{V}\left\{\because 1\mu ={10}^{-6},1\text{k}={\text{10}}^3\right\}\\ =\left(4\times {10}^{-3}\right)\left(3-0\right)\frac{\text{F}\cdot \text{V}}{\text{s}}\\ =\left(4\times {10}^{-3}\right)\left(3\right)\text{A}\left\{\because 1\text{A}=\frac{\text{1F}\cdot 1\text{V}}{\text{1s}}\right\}\end{array}$

Simplify the equation to find $i_2\left(t\right)$.

$\begin{array}{c}{i}_2\left(t\right)=\left(12\times {10}^{-3}\right)\text{A}\\ =12\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

For $3\text{s}<t<5\text{s}$:

Substitute $4\mu \text{F}$ for $C$ and $\left(0.75t^2-7.5t+23.25\right)\text{kV}$ for $v\left(t\right)$ in equation (10) to find $i_2\left(t\right)$.

$\begin{array}{c}{i}_2\left(t\right)=\left(4\mu \text{F}\right)\frac{d\left(\left(0.75t^2-7.5t+23.25\right)\text{kV}\right)}{dt}\\ =\left(4\times {10}^{-6}\right)\left({10}^3\right)\frac{d\left(0.75t^2-7.5t+23.25\right)}{dt}\text{F}\cdot \text{V}\left\{\because 1\mu ={10}^{-6},1\text{k}={\text{10}}^3\right\}\\ =\left(4\times {10}^{-3}\right)\left(1.5t-7.5+0\right)\frac{\text{F}\cdot \text{V}}{\text{s}}\\ =\left(4\times {10}^{-3}\right)\left(1.5t-7.5\right)\text{A}\left\{\because 1\text{A}=\frac{\text{1F}\cdot 1\text{V}}{\text{1s}}\right\}\end{array}$

Simplify the equation to find $i_2\left(t\right)$.

$\begin{array}{c}{i}_2\left(t\right)=\left(4\right)\left(1.5t-7.5\right)\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\\ =\left(6t-30\right)\text{mA}\end{array}$

Therefore, the expression of current through the capacitor $C_2$ is,

$i_2\left(t\right)=\{\begin{array}{l}12t\text{mA},0\text{s}<t<1\text{s}\\ 12\text{mA},1\text{s}<t<3\text{s}\\ \left(6t-30\right)\text{mA,}3\text{s}<t<5\text{s}\end{array}$

Conclusion:

Thus, the expression of voltage $v\left(t\right)$, current $i_1\left(t\right)$ and $i_2\left(t\right)$ is,

$v\left(t\right)=\{\begin{array}{l}1.5t^2\text{kV},0\text{s}<t<1\text{s}\\ \left(3t-1.5\right)\text{kV},1\text{s}<t<3\text{s}\\ \left(0.75t^2-7.5t+23.25\right)\text{kV,}3\text{s}<t<5\text{s}\end{array}$

$i_1\left(t\right)=\{\begin{array}{l}18t\text{mA},0\text{s}<t<1\text{s}\\ 18\text{mA},1\text{s}<t<3\text{s}\\ \left(9t-45\right)\text{mA,}3\text{s}<t<5\text{s}\end{array}$

$i_2\left(t\right)=\{\begin{array}{l}12t\text{mA},0\text{s}<t<1\text{s}\\ 12\text{mA},1\text{s}<t<3\text{s}\\ \left(6t-30\right)\text{mA,}3\text{s}<t<5\text{s}\end{array}$.