Chapter 4, Problem 4.15HP

The voltage across a generic element X has the waveform shown in Figure P4.15. For $0<t<10ms$ , determine and plot the current through X when it ¡s a:
a. $7\text{-}\Omega$ resistor.
b. $0.5\text{-}\mu F$ capacitor.
c. 7-mH inductor.

To determine

(a)

The current through the given element for the given voltage waveform.

To plot:

A graph of current through the element for $0<t<10\text{ms}$ .

Answer to Problem 4.15HP

The current through the given element for the given voltage waveform is

  $i_R\left(t\right)=\{\begin{array}{c}421.571t\text{A}0<t<5\text{ms}\\ 2.1428\text{A }5\text{ms}<t<10\text{ms}\end{array}$

A graph of current through the element for $0<t<10\text{ms}$ is shown in Figure 1.

Explanation of Solution

Given information:

Element is resistor and its value is $7\Omega$ .

The given voltage waveform is shown below.

  

Calculation:

Consider the given graph for the voltage.

For $0<t<5\text{ms}$, the equation of voltage will be

  $\begin{array}{c}v\left(t\right)-0=\frac{15-0}{\left(5-0\right)\times {10}^{-3}}\left(t-0\right)\\ v\left(t\right)=\frac{15}{5}\times {10}^{3}t\\ v\left(t\right)=3000t\end{array}$

For $5\text{ms}<t<10\text{ms}$, the voltage is constant and equal to 15 V.

So, voltage can be written as

  $v\left(t\right)=\{\begin{array}{c}3000t\text{V}0<t<5\text{ms}\\ 15\text{V }5\text{ms}<t<10\text{ms}\end{array}$

The given element is resistor. The current across it will be

  $i_R\left(t\right)=\frac{v_R\left(t\right)}{R}$

Voltage $v_R\left(t\right)$ will be

  $v_R\left(t\right)=v\left(t\right)=\{\begin{array}{c}3000t\text{V}0<t<5\text{ms}\\ 15\text{V }5\text{ms}<t<10\text{ms}\end{array}$

For $0<t<5\text{ms}$, the current will be

  $\begin{array}{c}{i}_R\left(t\right)=\frac{v_R\left(t\right)}{R}\\ i_R\left(t\right)=\frac{3000t}{7}\\ =428.571t\text{A}\end{array}$

For $5\text{ms}<t<10\text{ms}$, the current will be

  $\begin{array}{c}{i}_R\left(t\right)=\frac{v_R\left(t\right)}{R}\\ i_R\left(t\right)=\frac{15}{7}\\ =2.1428\text{A}\end{array}$

Current $i_R\left(t\right)$ will be

  $i_R\left(t\right)=\{\begin{array}{c}421.571t\text{A}0<t<5\text{ms}\\ 2.1428\text{A }5\text{ms}<t<10\text{ms}\end{array}$

A graph of current with respect to time is shown below.

  

Figure 1

To determine

(b)

The current through the given element for the given voltage waveform.

To plot:

A graph of current through the element for $0<t<10\text{ms}$ .

Answer to Problem 4.15HP

The current through the given element for the given voltage waveform is

  $i_C\left(t\right)=\{\begin{array}{c}1.5\text{mA}0<t<5\text{ms}\\ 0\text{mA }5\text{ms}<t<10\text{ms}\end{array}$

A graph of current through the element for $0<t<10\text{ms}$ is shown in Figure 2.

Explanation of Solution

Given information:

Element is capacitor and its value is $0.5\mu \text{F}$ .

The given voltage waveform is shown below.

  

Calculation:

Consider the given graph for the voltage.

For $0<t<5\text{ms}$, the equation of voltage will be

  $\begin{array}{c}v\left(t\right)-0=\frac{15-0}{\left(5-0\right)\times {10}^{-3}}\left(t-0\right)\\ v\left(t\right)=\frac{15}{5}\times {10}^{3}t\\ v\left(t\right)=3000t\end{array}$

For $5\text{ms}<t<10\text{ms}$, the voltage is constant and equal to 15 V.

So, voltage can be written as

  $v\left(t\right)=\{\begin{array}{c}3000t\text{V}0<t<5\text{ms}\\ 15\text{V }5\text{ms}<t<10\text{ms}\end{array}$

The given element is capacitor. The current across it will be

  $i_C\left(t\right)=C\frac{d\left(v_C\left(t\right)\right)}{dt}$

Voltage $v_C\left(t\right)$ will be

  $v_C\left(t\right)=v\left(t\right)=\{\begin{array}{c}3000t\text{V}0<t<5\text{ms}\\ 15\text{V }5\text{ms}<t<10\text{ms}\end{array}$

For $0<t<5\text{ms}$, the current will be

  $\begin{array}{c}{i}_C\left(t\right)=C\frac{d\left(v_C\left(t\right)\right)}{dt}\\ i_C\left(t\right)=\left(0.5\times {10}^{-6}\right)\frac{d\left(3000t\right)}{dt}\\ =\left(0.5\times {10}^{-6}\right)\left(3000\right)\\ =1500\times {10}^{-6}\text{A}\\ \text{=1}\text{.5}\times {\text{10}}^{-3}\text{A}\\ \text{=1}\text{.5}\text{mA}\end{array}$

For $5\text{ms}<t<10\text{ms}$, the current will be

  $\begin{array}{c}{i}_C\left(t\right)=\left(0.5\times {10}^{-6}\right)\frac{d\left(15\right)}{dt}\\ =\left(0.5\times {10}^{-6}\right)\left(0\right)\\ =0\text{mA}\end{array}$

Current $i_C\left(t\right)$ will be

  $i_C\left(t\right)=\{\begin{array}{c}1.5\text{mA}0<t<5\text{ms}\\ 0\text{mA }5\text{ms}<t<10\text{ms}\end{array}$

A graph of current with respect to time is shown below.

  

Figure 2

To determine

(c)

The current through the given element for the given voltage waveform.

To plot:

A graph of current through the element for $0<t<10\text{ms}$ .

Answer to Problem 4.15HP

The current through the given element for the given voltage waveform is

  $i_L\left(t\right)=\{\begin{array}{c}5.35714\text{A}0<t<5\text{ms}\\ 16.047114\text{A }5\text{ms}<t<10\text{ms}\end{array}$

A graph of current through the element for $0<t<10\text{ms}$ is shown in Figure 3.

Explanation of Solution

Given information:

Element is inductor and its value is $7\text{mH}$ .

The given voltage waveform is shown below.

  

Calculation:

Consider the given graph for the voltage.

For $0<t<5\text{ms}$, the equation of voltage will be

  $\begin{array}{c}v\left(t\right)-0=\frac{15-0}{\left(5-0\right)\times {10}^{-3}}\left(t-0\right)\\ v\left(t\right)=\frac{15}{5}\times {10}^{3}t\\ v\left(t\right)=3000t\end{array}$

For $5\text{ms}<t<10\text{ms}$, the voltage is constant and equal to 15 V.

So, voltage can be written as

  $v\left(t\right)=\{\begin{array}{c}3000t\text{V}0<t<5\text{ms}\\ 15\text{V }5\text{ms}<t<10\text{ms}\end{array}$

The given element is inductor. The current across it will be

  $i_L\left(t\right)=\frac{1}{L}{\displaystyle \underset{t_o}{\overset{t}{\int }}{v}_L\left(t\right)dt+i_L\left(t_o\right)}$

Voltage $v_L\left(t\right)$ will be

  $v_L\left(t\right)=v\left(t\right)=\{\begin{array}{c}3000t\text{V}0<t<5\text{ms}\\ 15\text{V }5\text{ms}<t<10\text{ms}\end{array}$

For $0<t<5\text{ms}$, the current will be

  $\begin{array}{c}{i}_L\left(t\right)=\frac{1}{L}{\displaystyle \underset{t_o}{\overset{t}{\int }}{v}_L\left(t\right)dt+i_L\left(t_o\right)}\\ i_L\left(t\right)=\frac{1}{7\times {10}^{-3}}{\displaystyle \underset{0}{\overset{5}{\int }}\left(3000t\right)dt+i_L\left(0\right)}\\ =\frac{1\times 3000}{7\times {10}^{-3}}{\left[\frac{t^2}{2}\right]}_0^{5\times {10}^{-3}}+0\\ =\frac{1\times 3000}{7\times {10}^{-3}}\left(\frac{25\times {10}^{-6}}{2}\right)\\ =\frac{1\times 3000\times 25\times {10}^{-6}}{7\times {10}^{-3}\times 2}\\ =\frac{75000\times {10}^{-6}}{14\times {10}^{-3}}\\ =5.35714\text{A}\end{array}$

For $5\text{ms}<t<10\text{ms}$, the current will be

  $\begin{array}{c}{i}_L\left(t\right)=\frac{1}{L}{\displaystyle \underset{t_o}{\overset{t}{\int }}{v}_L\left(t\right)dt+i_L\left(t_o\right)}\\ i_L\left(t\right)=\frac{1}{7\times {10}^{-3}}{\displaystyle \underset{5\times {10}^{-3}}{\overset{10\times {10}^{-3}}{\int }}\left(15\right)dt+i_L\left(5\text{ms}\right)}\\ =\frac{1\times 15}{7\times {10}^{-3}}{\left[t\right]}_{5\times {10}^{-3}}^{10\times {10}^{-3}}+5.35714\text{A}\\ =\frac{1\times 15}{7\times {10}^{-3}}\left(\left(10-5\right)\times {10}^{-3}\right)+5.35714\text{A}\\ \text{=}\left(\text{2}\text{.1428}\times {\text{10}}^3\times 5\times {10}^{-3}\right)+5.35714\text{A}\\ =10.714+5.35714\\ =16.07114\text{A}\end{array}$

Current $i_L\left(t\right)$ will be

  $i_L\left(t\right)=\{\begin{array}{c}5.35714\text{A}0<t<5\text{ms}\\ 16.047114\text{A }5\text{ms}<t<10\text{ms}\end{array}$

A graph of current with respect to time is shown below.

  

Figure 3