Chapter 4, Problem 4.13HP

To determine

The equation of a current through capacitor.

To plot:

The waveform of the current through the capacitor as a function of time.

To explain:

The effect on current due to discontinuities in the slope of the voltage waveform.

Answer to Problem 4.13HP

The equation for current through the capacitor for different time interval is given by,

  $v\left(t\right)=\{\begin{array}{l}510\text{mA}\text{for}\text{0}<t<\frac{2T}{3}\\ -1020\text{mA}\text{for}\frac{2T}{3}<t<T\\ 510\text{mA}\text{for}T<t<\frac{5T}{3}\\ -1020\text{mA}\text{for}\frac{5T}{3}<t<2T\\ 0\text{for}t\text{>}2T\end{array}$

Theplot for the waveform of the capacitor current is shown in Figure 3.

There are dissimilarities in the positive and the negative peak of the current because of the dissimilarities in voltage waveform.

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1.

 

The conversion from $\mu \text{s}$ to $\text{s}$ is given by,

  $1\mu \text{s}={10}^{-6}\text{s}$

The conversion from $\frac{80}{3}\mu \text{s}$ to $\text{s}$ is given by,

  $\frac{80}{3}\mu \text{s}=\frac{80}{3}\times {10}^{-6}\text{s}$

The conversion from $40\mu \text{s}$ to $\text{s}$ is given by,

  $40\mu \text{s}=40\times {10}^{-6}\text{s}$

Mark the values and draw the waveform of the current through the capacitor.

The required diagram is shown in Figure 2

 

From the above graph the voltage between the points $\left(0,0\right)$ and $\left(\frac{80\mu \text{s}}{3},20\right)$ is calculated as,

  $\begin{array}{c}v\left(t\right)=\left(\frac{0-20}{40\times {10}^{-6}-\frac{80\times {10}^{-6}}{3}}\right)\left(t-0\right)+0\\ =\left(7.5\times {10}^5\right)t\end{array}$

From the above graph the voltage between the points $\left(\frac{80\mu \text{s}}{3},20\right)$ and $\left(40\mu \text{s},0\right)$ is calculated as,

  $\begin{array}{c}v\left(t\right)=\left(\frac{0-20}{40\times {10}^{-6}-\frac{80\times {10}^{-6}}{3}}\right)\left(t-\frac{80\times {10}^{-6}}{3}\right)+20\\ =\left(-1.5\times {10}^6\right)t+40+20\\ =60-\left(1.5\times {10}^6\right)t\end{array}$

From the above graph the voltage between the points $\left(40\mu \text{s},0\right)$ and $\left(\frac{200\mu \text{s}}{3},20\right)$ is calculated as,

  $\begin{array}{c}v\left(t\right)=\left(\frac{20-0}{\frac{200\times {10}^{-6}}{3}-40\times {10}^{-6}}\right)\left(t-40\times {10}^{-6}\right)+20\\ =\left(7.5\times {10}^5\right)t-30\end{array}$

From the above graph, the voltage between the points $\left(\frac{200\mu \text{s}}{3},20\right)$ and $\left(80\mu \text{s},0\right)$ is calculated as,

  $\begin{array}{c}v\left(t\right)=\left(\frac{0-20}{80\times {10}^{-6}-\frac{200\times {10}^{-6}}{3}}\right)\left(t-\frac{200\times {10}^{-6}}{3}\right)+20\\ =\left(-1.5\times {10}^6\right)t+100+20\\ =120-\left(1.5\times {10}^6\right)t\end{array}$

The expression for the voltage for different time interval is,

  $v\left(t\right)=\{\begin{array}{l}\left(7.5\times {10}^5\right)t\text{for}\text{0}<t<\frac{2T}{3}\\ 60-\left(1.5\times {10}^6\right)t\text{for}\frac{2T}{3}<t<T\\ \left(7.5\times {10}^5\right)t-30\text{for}T<t<\frac{5T}{3}\\ 120-\left(1.5\times {10}^6\right)t\text{for}\frac{5T}{3}<t<2T\\ 0\text{for}t\text{>}2T\end{array}$

The conversion of $\text{nF}$ into $\text{F}$ is given by,

  $1\text{nF}={10}^{-6}\text{F}$

The conversion of $680\text{nF}$ into $\text{F}$ is given by,

  $680\text{nF}=680\times {10}^{-9}\text{F}$

The expression for the current through the capacitor is given by,

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

Substitute $680\times {10}^{-9}\text{F}$ for $C$ and $\left(7.5\times {10}^5\right)t$ for $v\left(t\right)$ in the above equation.

  $\begin{array}{c}i\left(t\right)=\left(680\times {10}^{-9}\text{F}\right)\frac{d\left[\left(7.5\times {10}^5\right)t\right]}{dt}\\ =\left(680\times {10}^{-9}\text{F}\right)\left(7.5\times {10}^5\right)\\ =510\times {10}^{-3}\text{A}\end{array}$

Substitute $680\times {10}^{-9}\text{F}$ for $C$ and $60-\left(1.5\times {10}^6\right)t$ for $v\left(t\right)$ in the above equation

  $\begin{array}{c}i\left(t\right)=\left(680\times {10}^{-9}\text{F}\right)\frac{d\left[60-\left(1.5\times {10}^6\right)t\right]}{dt}\\ =-\left(680\times {10}^{-9}\text{F}\right)\left(1.5\times {10}^6\right)\\ =-1020\times {10}^{-3}\text{A}\end{array}$

Substitute $680\times {10}^{-9}\text{F}$ for $C$ and $\left(7.5\times {10}^5\right)t-30$ for $v\left(t\right)$ in the above equation.

  $\begin{array}{c}i\left(t\right)=\left(680\times {10}^{-9}\text{F}\right)\frac{d\left[\left(7.5\times {10}^5\right)t-30\right]}{dt}\\ =\left(680\times {10}^{-9}\text{F}\right)\left(7.5\times {10}^5\right)\\ =510\times {10}^{-3}\text{A}\end{array}$

Substitute $680\times {10}^{-9}\text{F}$ for $C$ and $120-\left(1.5\times {10}^6\right)t$ for $v\left(t\right)$ in the above equation

  $\begin{array}{c}i\left(t\right)=\left(680\times {10}^{-9}\text{F}\right)\frac{d\left[120-\left(1.5\times {10}^6\right)t\right]}{dt}\\ =-\left(680\times {10}^{-9}\text{F}\right)\left(1.5\times {10}^6\right)\\ =-1020\times {10}^{-3}\text{A}\end{array}$

The conversion from $\text{A}$ into $\text{mA}$ is given by,

  $1\text{A}={10}^3\text{mA}$

The conversion from $510\times {10}^{-3}\text{A}$ into $\text{mA}$ is given by,

  $510\times {10}^{-3}\text{A}=510\text{mA}$

The conversion from $1020\times {10}^{-6}\text{A}$ into $\text{mA}$ is given by,

  $1020\times {10}^{-3}\text{A}=1020\text{mA}$

The equation for current through the capacitor for different time interval is given by,

  $v\left(t\right)=\{\begin{array}{l}510\text{mA}\text{for}\text{0}<t<\frac{2T}{3}\\ -1020\text{mA}\text{for}\frac{2T}{3}<t<T\\ 510\text{mA}\text{for}T<t<\frac{5T}{3}\\ -1020\text{mA}\text{for}\frac{5T}{3}<t<2T\\ 0\text{for}t\text{>}2T\end{array}$

The waveform of the current through the capacitor is shown below.

The required diagram is shown in Figure 3

 

From the above Figure, it is clear that positive and the negative peak amplitudes of the current through the capacitor are differentbecause of the discontinuities in slope in the voltage waveform.

Conclusion:

Therefore, the plot for the waveform of the capacitor current is shown in Figure 3, and there are dissimilarities in the positive and the negative peak of the current because of the dissimilarities in voltage waveform.