Chapter 6, Problem 6P

The voltage waveform in Fig. 6.46 is applied across a 55-μF capacitor. Draw the current waveform through it.

Figure 6.46

For Prob. 6.6.

To determine

Find the current waveform.

Explanation of Solution

Given data:

The value of the capacitor $\left(C\right)$ is $55\mu \text{F}$.

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 6.46 in the textbook.

From the given graph, substitute $t$ for $x$ and $v\left(t\right)$ for $y$ in equation (1).

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

Write the expression to calculate the current through the inductor.

$i\left(t\right)=C\frac{dv\left(t\right)}{dt}$ (3)

Here,

$C$ is the value of the capacitor, and

$\frac{dv\left(t\right)}{dt}$ is the rate of change of voltage with time.

Calculation:

The given voltage waveform is redrawn as Figure 1.

 

Refer to Figure 1, split up the time period as five divisions such as $0\text{ms}<t<2\text{ms}$, $2\text{ms}<t<4\text{ms}$, $4\text{ms}<t<8\text{ms}$, $8\text{ms}<t<10\text{ms}$ and $10\text{ms}<t<12\text{ms}$ to find the respective voltage value.

Case (i): $0\text{ms}<t<2\text{ms}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(0\text{ms},\text{0}\text{V}\right)$ and $\left(2\text{ms},\text{10}\text{V}\right)$.

Substitute $0\text{ms}$ for $x_1$, $\text{0}\text{V}$ for $y_1$, $2\text{ms}$ for $x_2$ and $\text{10}\text{V}$ for $y_2$ in equation (2) to find $v\left(t\right)$.

$\left(v\left(t\right)-\text{0}\text{V}\right)=\frac{\text{10}\text{V}-\text{0}\text{V}}{2\text{ms}-0\text{ms}}\left(t-0\text{ms}\right)$

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

$\begin{array}{c}v\left(t\right)=\frac{\text{10}\text{V}}{2\text{ms}}\left(t\right)\\ =\frac{\text{10}}{2\times {10}^{-3}}t\text{V}\left\{\because 1\text{m}={\text{10}}^{-3}\right\}\\ =\left(5000t\right)\text{V}\end{array}$

Case (ii): $2\text{ms}<t<4\text{ms}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(4\text{ms},1\text{0}\text{V}\right)$ and $\left(2\text{ms},\text{10}\text{V}\right)$.

Substitute $4\text{ms}$ for $x_1$, $1\text{0}\text{V}$ for $y_1$, $2\text{ms}$ for $x_2$ and $\text{10}\text{V}$ for $y_2$ in equation (2) to find $v\left(t\right)$.

$\begin{array}{l}v\left(t\right)-1\text{0}\text{V}=\frac{\text{10}\text{V}-1\text{0}\text{V}}{2\text{ms}-4\text{ms}}\left(t-4\text{ms}\right)\\ v\left(t\right)-1\text{0}\text{V}=\frac{0\text{V}}{-2\text{ms}}\left(t-4\text{ms}\right)\\ v\left(t\right)-1\text{0}\text{V}=0\end{array}$

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

$v\left(t\right)=1\text{0}\text{V}$

Case (iii): $4\text{ms}<t<8\text{ms}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(8\text{ms},-10\text{V}\right)$ and $\left(4\text{ms},1\text{0}\text{V}\right)$.

Substitute $8\text{ms}$ for $x_1$, $-\text{10}\text{V}$ for $y_1$, $4\text{ms}$ for $x_2$ and $\text{10}\text{V}$ for $y_2$ in equation (2) to find $v\left(t\right)$.

$\begin{array}{l}\left(v\left(t\right)-\left(-\text{10}\text{V}\right)\right)=\frac{\text{10}\text{V}-\left(-\text{10}\text{V}\right)}{4\text{ms}-8\text{ms}}\left(t-8\text{ms}\right)\\ \left(v\left(t\right)+10\text{V}\right)=\frac{\text{10}\text{V}+\text{10}\text{V}}{-4\text{ms}}\left(t-8\text{ms}\right)\\ \left(v\left(t\right)+10\text{V}\right)=-\frac{20\text{V}}{4\text{ms}}\left(t-8\text{ms}\right)\end{array}$

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

$\begin{array}{c}v\left(t\right)=\left(-\frac{20\text{V}}{4\text{ms}}\left(t-8\text{ms}\right)-10\text{V}\right)\\ =\left(-\left(\frac{20}{4\times {10}^{-3}}\right)\left(t-\left(8\times {10}^{-3}\right)\right)-10\right)\text{V}\left\{\because 1\text{m}={\text{10}}^{-3}\right\}\\ =\left(-5000t+40-10\right)\text{V}\\ =\left(-5000t+30\right)\text{V}\end{array}$

Case (iv): $8\text{ms}<t<10\text{ms}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(8\text{ms},-1\text{0}\text{V}\right)$ and $\left(10\text{ms},-\text{10}\text{V}\right)$.

Substitute $8\text{ms}$ for $x_1$, $-1\text{0}\text{V}$ for $y_1$, $10\text{ms}$ for $x_2$ and $-\text{10}\text{V}$ for $y_2$ in equation (2) to find $v\left(t\right)$.

$\begin{array}{l}v\left(t\right)-\left(-1\text{0}\text{V}\right)=\frac{\left(-\text{10}\text{V}\right)-\left(-1\text{0}\text{V}\right)}{10\text{ms}-8\text{ms}}\left(t-8\text{ms}\right)\\ v\left(t\right)+1\text{0}\text{V}=\frac{-10\text{V+10}\text{V}}{2\text{ms}}\left(t-8\text{ms}\right)\\ v\left(t\right)+1\text{0}\text{V}=\frac{0\text{V}}{2\text{ms}}\left(t-8\text{ms}\right)\\ v\left(t\right)+1\text{0}\text{V}=0\end{array}$

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

$v\left(t\right)=-1\text{0}\text{V}$

Case (v): $10\text{ms}<t<12\text{ms}$

The two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ are $\left(12\text{ms},0\text{V}\right)$ and $\left(10\text{ms},-1\text{0}\text{V}\right)$.

Substitute $12\text{ms}$ for $x_1$, $\text{0}\text{V}$ for $y_1$, $10\text{ms}$ for $x_2$ and $-\text{10}\text{V}$ for $y_2$ in equation (2) to find $v\left(t\right)$.

$\left(v\left(t\right)-\text{0}\text{V}\right)=\frac{-\text{10}\text{V}-\text{0}\text{V}}{10\text{ms}-12\text{ms}}\left(t-12\text{ms}\right)$

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

$\begin{array}{c}v\left(t\right)=\frac{-\text{10}\text{V}}{-2\text{ms}}\left(t-12\text{ms}\right)\\ =\left(\frac{\text{10}}{2\times {10}^{-3}}\left(t-\left(12\times {10}^{-3}\right)\right)\right)\text{V}\left\{\because 1\text{m}={\text{10}}^{-3}\right\}\\ =\left(5000t-60\right)\text{V}\end{array}$

Therefore, the voltage function of the signal in Figure 1 is,

$v\left(t\right)=\{\begin{array}{l}\left(5000t\right)\text{V},0\text{ms}<t<2\text{ms}\\ 1\text{0}\text{V},2\text{ms}<t<4\text{ms}\\ \left(-5000t+30\right)\text{V},4\text{ms}<t<8\text{ms}\\ -1\text{0}\text{V,}8\text{ms}<t<10\text{ms}\\ \left(5000t-60\right)\text{V,}10\text{ms}<t<12\text{ms}\end{array}$

For $0\text{ms}<t<2\text{ms}$:

Substitute $55\mu \text{F}$ for $C$ and $5000t\text{V}$ for $v\left(t\right)$ in equation (3) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left(55\mu \text{F}\right)\frac{d}{dt}\left(5000t\text{V}\right)\\ =\left(55\times {10}^{-6}\right)\left(5000\right)\frac{d}{dt}\left(t\right)\text{V}\cdot \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\\ =\left(55\times {10}^{-6}\right)\left(5000\right)\left(1\right)\frac{\text{V}\cdot \text{F}}{\text{s}}\\ =0.275\text{A}\left\{\because 1\text{A}=\frac{\text{1V}\cdot 1\text{F}}{\text{1s}}\right\}\end{array}$

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

$\begin{array}{c}i\left(t\right)=0.275\times {10}^3\times {10}^{-3}\text{A}\\ =275\text{mA}\left\{\because 1\text{m}={\text{10}}^3\right\}\end{array}$

For $2\text{ms}<t<4\text{ms}$:

Substitute $55\mu \text{F}$ for $C$ and $1\text{0}\text{V}$ for $v\left(t\right)$ in equation (3) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left(55\mu \text{F}\right)\frac{d}{dt}\left(1\text{0}\text{V}\right)\\ =\left(55\times {10}^{-6}\right)\frac{d}{dt}\left(10\right)\text{V}\cdot \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\\ =\left(55\times {10}^{-6}\right)\left(0\right)\frac{\text{V}\cdot \text{F}}{\text{s}}\\ =0\text{A}\left\{\because 1\text{A}=\frac{\text{1V}\cdot 1\text{F}}{\text{1s}}\right\}\end{array}$

For $4\text{ms}<t<8\text{ms}$:

Substitute $55\mu \text{F}$ for $C$ and $\left(-5000t+30\right)\text{V}$ for $v\left(t\right)$ in equation (3) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left(55\mu \text{F}\right)\frac{d}{dt}\left(-5000t+30\right)\text{V}\\ =\left(55\times {10}^{-6}\right)\frac{d}{dt}\left(-5000t+30\right)\text{V}\cdot \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\\ =\left(55\times {10}^{-6}\right)\left(-5000\left(1\right)+0\right)\frac{\text{V}\cdot \text{F}}{\text{s}}\\ =-0.275\text{A}\left\{\because 1\text{A}=\frac{\text{1V}\cdot 1\text{F}}{\text{1s}}\right\}\end{array}$

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

$\begin{array}{c}i\left(t\right)=-0.275\times {10}^3\times {10}^{-3}\text{A}\\ =-275\text{mA}\left\{\because 1\text{m}={\text{10}}^3\right\}\end{array}$

For $8\text{ms}<t<10\text{ms}$:

Substitute $55\mu \text{F}$ for $C$ and $-1\text{0}\text{V}$ for $v\left(t\right)$ in equation (3) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left(55\mu \text{F}\right)\frac{d}{dt}\left(-1\text{0}\text{V}\right)\\ =\left(55\times {10}^{-6}\right)\frac{d}{dt}\left(-10\right)\text{V}\cdot \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\\ =\left(55\times {10}^{-6}\right)\left(0\right)\frac{\text{V}\cdot \text{F}}{\text{s}}\\ =0\text{A}\left\{\because 1\text{A}=\frac{\text{1V}\cdot 1\text{F}}{\text{1s}}\right\}\end{array}$

For $10\text{ms}<t<12\text{ms}$:

Substitute $55\mu \text{F}$ for $C$ and $\left(5000t-60\right)\text{V}$ for $v\left(t\right)$ in equation (3) to find $i\left(t\right)$.

$\begin{array}{c}i\left(t\right)=\left(55\mu \text{F}\right)\frac{d}{dt}\left(5000t-60\right)\text{V}\\ =\left(55\times {10}^{-6}\right)\frac{d}{dt}\left(5000t-60\right)\text{V}\cdot \text{F}\left\{\because 1\mu ={10}^{-6}\right\}\\ =\left(55\times {10}^{-6}\right)\left(5000\left(1\right)-0\right)\frac{\text{V}\cdot \text{F}}{\text{s}}\\ =0.275\text{A}\left\{\because 1\text{A}=\frac{\text{1V}\cdot 1\text{F}}{\text{1s}}\right\}\end{array}$

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

$\begin{array}{c}i\left(t\right)=0.275\times {10}^3\times {10}^{-3}\text{A}\\ =275\text{mA}\left\{\because 1\text{m}={\text{10}}^3\right\}\end{array}$

Therefore, the current function of the signal in Figure 1 is,

$i\left(t\right)=\{\begin{array}{l}275\text{mA},0\text{ms}<t<2\text{ms}\\ 0\text{A},2\text{ms}<t<4\text{ms}\\ -275\text{mA},4\text{ms}<t<8\text{ms}\\ 0\text{A,}8\text{ms}<t<10\text{ms}\\ 275\text{mA,}10\text{ms}<t<12\text{ms}\end{array}$

From the current expression $i\left(t\right)$, the current waveform is drawn in Figure 2.

Conclusion:

Thus, the current equation is found and its respective waveform is drawn.