Chapter 4, Problem 4.74HP

To determine

The value of node voltages $v_a\left(t\right)$ and $v_b\left(t\right)$ .

Answer to Problem 4.74HP

The value of voltage $v_a\left(t\right)$ is $7.56\cos \left(\omega t+28.12°\right)\text{V}$ and the value of the voltage $v_b\left(t\right)$ is $7.89\cos \left(300t+15°\right)$.

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

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

  $1\mu \text{F}={10}^{-6}\text{F}$

The conversion from $300\mu \text{F}$ into $\text{F}$ is given by

  $300\mu \text{F}=300\times {10}^{-6}\text{F}$

The conversion from $1\text{mH}$ into $\text{H}$ is given by

  $1\text{mH}={10}^{-3}\text{H}$

The conversion from $300\text{mH}$ into $\text{H}$ is given by

  $300\text{mH}=300\times {10}^{-3}\text{H}$

The expression for the source voltage is,

  $v\left(t\right)=7\cos \left(300t+\frac{\pi }{4}\right)\text{V}$ ...... (1)

The general for the time-dependent expression of voltage is given by

  $v\left(t\right)=V_m\cos \left(\omega t+\varphi \right)$ ...... (2)

From above and from equation (1) the value of angular frequency is given by

  $\omega =300\text{rad}/\text{s}$

The general form for the polar form of the voltage is given by

  $V=V_0\angle \varphi °$

From equation (1) and above, the polar form of the supply voltage is given by

  $\begin{array}{c}V=7\angle \frac{\pi }{4}°\\ =7\angle 45°\end{array}$

The expression for the source current is given by

  $i\left(t\right)=2\cos \left(300t\right)\text{A}$ ...... (3)

The general expression for the phasor form of the current is given by

  $I=I_0\angle \varphi °$

From equation (3) and equation (1), the phasor form of the current is given by

  $I=2\angle 0°\text{A}$

The expression to calculate the inductive impedance of the inductor is given by

  $Z_L=j\omega L$

Substitute $300\pi \text{rad}/\text{s}$ for $\omega$ and $300\times {10}^{-3}\text{H}$ for $L$ in the above equation.

  $\begin{array}{c}{Z}_L=j\left(300\text{rad}/\text{s}\right)\left(300\times {10}^{-3}\text{H}\right)\\ =j90\Omega \end{array}$

The expression to calculate the capacitive reactance is given by,

  $Z_C=\frac{1}{j\omega C}$

Substitute $300\pi \text{rad}/\text{s}$ for $\omega$ and $300\times {10}^{-6}\text{F}$ for $C$ in the above equation.

  $\begin{array}{c}{Z}_C=\frac{1}{j\left(300\pi \text{rad}/\text{s}\right)\left(300\times {10}^{-6}\text{F}\right)}\\ =-j11.11\Omega \end{array}$

Mark the values and redraw the circuit for the phasor form.

The required diagram is shown in Figure 2

  

Apply KCL at node $V_a$ .

  $\begin{array}{l}\frac{V_b-V_a}{3}+\frac{V_b-V_a}{-j11.11}+\frac{V_b}{5}-2\angle 0°=0\\ 0.333V_b-0.333V_a+j0.09V_b-j0.09V_a+0.2V_b=2\angle 0°\\ -\left(0.333+j0.09\right)V_a+\left(0.533-j0.09\right)V_b=2\\ \left(0.533+j0.09\right)V_b=2+\left(0.333+j0.09\right)V_a\end{array}$

Solve further as

  $V_b=\frac{2+\left(0.333+j0.09\right)V_a}{\left(0.533+j0.09\right)}$ ...... (4)

Apply KCL at node $V_a$ .

  $\begin{array}{l}\frac{V_a-7\angle 45°}{4}+\frac{V_a-V_b}{3}+\frac{V_a-V_b}{-j11.11}+\frac{V_a}{j90}=0\\ 0.25V_a-1.75\angle 45°+0.333V_a-0.333V_b+j0.09V_a-j0.09V_b-j0.011V_a=0\\ \left(0.583+j0.079\right)V_a-\left(0.333+j0.09\right)V_b=1.237+j1.237\end{array}$

Substitute $\frac{2+\left(0.333+j0.09\right)V_a}{\left(0.533+j0.09\right)}$ for $V_b$ in the above equation.

  $\begin{array}{l}\left(0.583+j0.079\right)V_a-\left(0.333+j0.09\right)\left(\frac{2+\left(0.333+j0.09\right)V_a}{\left(0.533+j0.09\right)}\right)=1.237+j1.237\\ \left(0.583+j0.079\right)V_a-\left(0.635+j0.06\right)\left(2+\left(0.333+j0.09V_a\right)\right)=1.237+j1.237\\ \left(0.377+j0.002\right)V_a=2.507+j1.357\\ V_a=\frac{2.507+j1.357}{0.377+j0.002}\end{array}$

Solve further as

  $\begin{array}{c}{V}_a=6.67+j3.56\text{V}\\ =\sqrt{{6.67}^2+{3.56}^2}{\tan }^{-1}\left(\frac{3.56}{6.67}\right)\\ =7.56\angle 28.12°\text{V}\end{array}$

From above and from equation (2), the time-dependent form of the node voltage $V_a$ is,

  $v_a\left(t\right)=7.56\cos \left(\omega t+28.12°\right)\text{V}$

Substitute $300\text{rad}/\text{s}$ for $\omega$ in the above equation.

  $v_a\left(t\right)=7.56\cos \left(300t+28.12°\right)\text{V}$

Substitute $6.67+j3.56\text{V}$ for $V_a$ in equation (4).

  $\begin{array}{c}{V}_b=\frac{2+\left(0.333+j0.09\right)\left(6.67+j3.56\text{V}\right)}{\left(0.533+j0.09\right)}\\ =\frac{2+1.88+j1.775}{\left(0.533+j0.09\right)}\\ =7.62+j2.042\text{V}\\ \text{=}\sqrt{{7.62}^2+{2.042}^2}{\tan }^{-1}\frac{2.042}{7.62}\end{array}$

Solve further as

  $V_b=7.89\angle 15°\text{V}$

From above and from equation (2), the time-dependent form of the node voltage $V_b$ is,

  $v_b\left(t\right)=7.89\cos \left(\omega t+15°\right)$

Substitute $300\text{rad}/\text{s}$ for $\omega$ in the above equation.

  $v_b\left(t\right)=7.89\cos \left(300t+15°\right)$

Conclusion:

Therefore, the value of voltage $v_a\left(t\right)$ is $7.56\cos \left(\omega t+28.12°\right)\text{V}$ and the value of the voltage $v_b\left(t\right)$ is $7.89\cos \left(300t+15°\right)$ .