Chapter 5, Problem 5.34HP

For $t<0$ , the circuit shown in Figure P5.34 is at steady state. The switch is thrown at $t=0$ . Assume:
$\begin{array}{l}{V}_{s1}=17V{V}_{s2}=11V\\ R_1=14\Omega R_2=13\Omega \\ R_3=14\Omega C=70nF\end{array}$
Determine the
a. Current $i_C$ through the capacitor for $t>0$ .
b. Voltage $v_3$ across $R_3$ for $t>0$ .
c. lime required for $i_C$ and $v_3$ to change by 98 percent of their initial values at $t=0^{+}$.

To determine

(a)

The value of the current $i_C$ for time $t>0$ .

Answer to Problem 5.34HP

The value of the current $i_C$ is $-0.575\times {10}^{-3}\left(e^{-714.29t}\right)\text{A}$ for time $t>0$ .

Explanation of Solution

Calculation:

The conversion from $\text{k}\Omega$ into $\Omega$ is given by,

  $\text{1}\text{k}\Omega ={10}^3\Omega$

The conversion from $14\text{k}\Omega$ into $\Omega$ is given by,

  $14\text{k}\Omega =14\times {10}^3\Omega$

The conversion from $13\text{k}\Omega$ into $\Omega$ is given by,

  $13\text{k}\Omega =13\times {10}^3\Omega$

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

  $\text{1}\text{nF}={10}^{-9}\text{F}$

The conversion from $\text{70}\text{nF}$ into $\text{F}$ is given by,

  $\text{70}\text{nF}=70\times {10}^{-9}\text{F}$

The given diagram is shown in Figure 1

  

For time $t<0$ the switch is connected to the point $1$ and the circuit is in steady state, thus the capacitor acts as an open circuit.

The required diagram is shown in Figure 2

  

From above circuit, the value of the voltage across the capacitor terminals for time $t=0^{-}$ is given by,

  $v_C\left(0^{-}\right)=V_{S1}$

Substitute $17\text{V}$ for $V_{S1}$ in the above circuit. $V_{S1}$

  $v_C\left(0^{-}\right)=17\text{V}$

From the above circuit,the expression for the value of the current through the capacitor for time $t=0^{-}$ is given by,

  $i_C\left(0^{-}\right)=0\text{A}$

For time $t=0^{+}$, the switch is moved to position 2 and the voltage across the capacitor for the time is given by,

  $v_C\left(0^{+}\right)=v_C\left(0^{-}\right)$

Mark the values and redraw the circuit for the time $t>0$ .

The required diagram is shown in Figure 3

  

Apply KVL at the node $v_3$ .

  $\begin{array}{l}\frac{v_3\left(0^{+}\right)-11}{14}+\frac{v_3\left(0^{+}\right)-17}{13}+\frac{v_3\left(0^{+}\right)}{14}=0\\ v_3\left(0^{+}\right)\left[\frac{20}{91}\right]=\frac{381}{182}\\ v_3\left(0^{+}\right)=9.525\text{V}\end{array}$

The expression for the current through the capacitor for time $t=0^{+}$ is given by,

  $i_C\left(0^{+}\right)=\frac{v_3\left(0^{+}\right)-17\text{V}}{13\text{k}\Omega }$

Substitute $9.525\text{V}$ for $v_3\left(0^{+}\right)$ in the above equation.

  $\begin{array}{c}{i}_C\left(0^{+}\right)=\frac{9.525\text{V}-17\text{V}}{13\times {10}^3\Omega }\\ =-0.575\times {10}^{-3}\text{A}\end{array}$

For time $t=\infty$, the capacitor is open circuited.

The required diagram is shown in Figure 4

  

From the above circuit, the expression for the value of the current through the for the time $t=\infty$ is given by,

  $i_C\left(\infty \right)=0\text{A}$

The expression for the voltage across the capacitor for the time $t=\infty$ is given by,

  $v_3\left(\infty \right)=V_{S2}\left(\frac{R_3}{R_3+R_1}\right)$

Substitute $11\text{V}$ for $V_{S2}$, $14\text{k}\Omega$ for $R_1$ and $14\text{k}\Omega$ for $R_3$ in the above equation.

  $\begin{array}{c}{v}_3\left(\infty \right)=\left(11\text{V}\right)\left(\frac{14\text{k}\Omega }{14\text{k}\Omega +14\text{k}\Omega }\right)\\ =5.5\text{V}\end{array}$

To calculate the Thevenin equivalent resistance for the circuit open circuit the capacitor terminals, short circuit the voltage source and redraw the circuit.

The required diagram is shown in Figure 5

  

From above figure the expression to calculate the time constant of the circuit is given by,

  $\tau =\left[\frac{R_1{R}_3}{R_1+R_3}+R_2\right]C$

Substitute $14\text{k}\Omega$ for $R_1$, $13\text{k}\Omega$ for $R_2$, $14\text{k}\Omega$ for $R_3$ and $70\times {10}^{-9}\text{F}$ for $C$ in the above equation.

  $\begin{array}{c}\tau =\left[\frac{\left(14\text{k}\Omega \right)\left(14\text{k}\Omega \right)}{\left(14\text{k}\Omega \right)+\left(14\text{k}\Omega \right)}+13\text{k}\Omega \right]\left(70\times {10}^{-9}\text{F}\right)\\ =\left[\frac{\left(14\times {10}^3\Omega \right)\left(14\times {10}^3\Omega \right)}{\left(14\times {10}^3\Omega \right)+\left(14\times {10}^3\Omega \right)}+13\times {10}^3\Omega \right]\left(70\times {10}^{-9}\text{F}\right)\\ =\left(7\times {10}^3\Omega +13\times {10}^3\Omega \right)\left(70\times {10}^{-9}\text{F}\right)\\ =1.4\times {10}^{-3}\text{s}\end{array}$

The expression for the complete solution for the current $i_L$ is given by,

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

Substitute $0\text{A}$ for $i_L\left(\infty \right)$, $-0.575\times {10}^{-3}\text{A}$ for $i_L\left(0^{+}\right)$ and $1.4\times {10}^{-3}\text{s}$ for $\tau$ in the above equation.

  $\begin{array}{c}{i}_L=0+\left[-0.575\times {10}^{-3}\text{A}-0\right]e^{\frac{-t}{1.4\times {10}^{-3}\text{s}}}\\ =-0.575\times {10}^{-3}\left(e^{-714.29t}\right)\text{A}\end{array}$

Conclusion:

Therefore, thevalue of the current $i_C$ is $-0.575\times {10}^{-3}\left(e^{-714.29t}\right)\text{A}$ for time $t>0$ .

To determine

(b)

The value of voltage $v_3$ across $R_3$ for $t>0$ .

Answer to Problem 5.34HP

Thevalue of the voltage $v_3$ for $t>0$ is $9.525\text{V}$ .

Explanation of Solution

Calculation:

The expression for the value of the voltage $v_3$ for $t>0$ is given by,

  $v_3\left(0^{+}\right)=9.525\text{V}$

Conclusion:

Therefore, the value of the voltage $v_3$ for $t>0$ is $9.525\text{V}$ .

To determine

(c)

The time required for $i_C$ and $v_3$ to change by $98\%$ of their initial values at $t=0^{+}$ .

Answer to Problem 5.34HP

Thetime required for $i_C$ to change $98\%$ of its initial value is $0.2828\times {10}^{-6}\text{F}$ and the time required by the voltage $v_3$ to change by $98\%$ is $0.68\times {10}^{-6}\text{sec}$.

Explanation of Solution

Calculation:

The expression for the current through the capacitor for time $t>0$ is given by,

  $i_C=-0.575\times {10}^{-3}{e}^{-714.29t}\text{A}$

Substitute $98\%\left(i_C\left(0^{+}\right)\right)$ for $i_C$ in the above equation.

  $98\%\left(i_C\left(0^{+}\right)\right)=-0.575\times {10}^{-3}{e}^{-714.29t}\text{A}$

Substitute $-0.575\times {10}^{-3}\text{A}$ for $i_C\left(0^{+}\right)$ in the above equation.

  $\begin{array}{l}\left[98\%\right]\left(-0.575\times {10}^{-3}\text{A}\right)=-0.575\times {10}^{-3}{e}^{-714.29t}\text{A}\\ e^{-714.29t}=0.98\\ -\mathrm{7.14.29}t=-0.0202\\ t=0.2828\times {10}^{-6}\text{F}\end{array}$

The expression for the current $v_3$ for time $t>0$ is given by,

  $v_3=v_3\left(\infty \right)+\left[v_3\left(0^{+}\right)-v_3\left(\infty \right)\right]e^{\frac{-t}{\tau }}$

Substitute $9.525\text{V}$ for $v_3\left(0^{+}\right)$, $5.5\text{V}$ for $v_3\left(\infty \right)$ and $1.4\times {10}^{-3}\text{sec}$ for $\tau$ in the above equation.

  $\begin{array}{c}{v}_3=5.5\text{V}+\left[9.525\text{V}-5.5\text{V}\right]e^{\frac{-t}{1.4\times {10}^{-3}\text{sec}}}\\ =5.5+4.025e^{-\mathrm{7.14.29}t}\end{array}$

Substitute $\left[98\%\right]\left[v_3\left(0^{+}\right)\right]$ for $v_3$ in the above equation

  $\left[98\%\right]\left[v_3\left(0^{+}\right)\right]=5.5+4.025e^{-\mathrm{7.14.29}t}$

Substitute $5.5+4.025e^{-\mathrm{7.14.29}t}$ for $v_3\left(0^{+}\right)$ in the above equation.

  $\begin{array}{l}\left[98\%\right]\left[5.5+4.025e^{-\mathrm{7.14.29}t}\right]=5.5+4.025e^{-\mathrm{7.14.29}t}\\ e^{-\mathrm{7.14.29}t}=0.98\\ -714.29t=-0.04856\\ t=0.68\times {10}^{-6}\text{sec}\end{array}$

Conclusion:

Therefore, the time required for $i_C$ to change $98\%$ of its initial value is $0.2828\times {10}^{-6}\text{F}$ and the time required by the voltage $v_3$ to change by $98\%$ is $0.68\times {10}^{-6}\text{sec}$ .