Chapter 4, Problem 4.76HP

Find the Thévenin equivalent network seen by the capacitor C in Figure P4.76. Use the result and voltage division to determine $v_c\left(t\right)$ . Assume:
$\begin{array}{l}{v}_c\left(t\right)=\cos \left(300t\right)V\\ i\left(t\right)=2\cos \left(300t\right)A\\ R_1=8\Omega R_2=8\Omega \\ L=3\mu HC=5\mu F\end{array}$

To determine

The Thevenin equivalent of the network seen by the capacitor $C$ and the value of the voltage $v_C\left(t\right)$ .

Answer to Problem 4.76HP

The Thevenin equivalent of the network seen by the capacitor $C$ is shown in Figure 5. The value of the voltage $v_C\left(t\right)$ is $8.5\cos \left(300t-0.35°\right)$.

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 $5\mu \text{F}$ into $\text{F}$ is given by,

  $5\mu \text{F}=5\times {10}^{-6}\text{F}$

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

  $1\mu \text{H}={10}^{-6}\text{H}$

The conversion from $3\mu \text{H}$ into $\text{H}$ is given by,

  $3\mu \text{H}=3\times {10}^{-6}\text{H}$

The general form for the expression of voltage is,

  $v\left(t\right)=V_m\cos \left(\omega t+\varphi \right)$ ....... (1)

The given expression for voltage is given by,

  $v\left(t\right)=\cos 300t\text{V}$

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

  $\omega =300\text{rad}/\text{s}$

The phasor form of the source voltage is given by,

  $V=V_0\angle \varphi °$

Substitute $1\text{V}$ for $V_0$ and $0$ for $\varphi$ in the above equation.

  $V=1\angle 0°$

The given expression for current is given by,

  $i\left(t\right)=3\cos \left(300\pi t\right)$ ....... (2)

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

  $I=I_0\angle \varphi °$

From above and from equation (2) the expression for the phasor form of the current is given by,

  $I=3\angle 0°\text{A}$

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

  $Z_L=j\omega L$

Substitute $300\text{rad}/\text{s}$ for $\omega$ and $3\times {10}^{-6}\text{H}$ for $L$ in the above equation .

  $\begin{array}{c}{Z}_L=j\left(300\text{rad}/\text{s}\right)\left(3\times {10}^{-6}\text{H}\right)\\ =j\left(9\times {10}^{-4}\right)\Omega \end{array}$

The expression to calculate the capacitive reactance is given by,

  $Z_C=\frac{1}{j\omega C}$

Substitute $300\text{rad}/\text{s}$ for $\omega$ and $5\times {10}^{-6}\text{F}$ for $C$ in the above equation.

  $\begin{array}{c}{Z}_C=\frac{1}{j\left(300\pi \text{rad}/\text{s}\right)\left(5\times {10}^{-6}\text{F}\right)}\\ =-j666.67\Omega \end{array}$

Mark the values and redraw the circuit for the frequency domain.

The required diagram is shown in Figure 2

  

To obtain the Thevenin equivalent of the circuit, open circuit the capacitor and redraw the circuit.

The required diagram is shown in Figure 3

  

From above figure the value of current $I_2$ is given by,

  $I_2=I$

Substitute $2\angle 0°\text{A}$ for $I$ in the above equation.

  $I_2=2\angle 0°\text{A}$

Apply KVL to the first loop.

  $\begin{array}{l}-1\angle 0°+8I_1+\left[j\left(9\times {10}^{-4}\right)\right]I_1+8\left(I_1+I_2\right)=0\\ \left[16+j\left(9\times {10}^{-4}\right)I_1\right]+8I_2=1\angle 0°\end{array}$

Substitute $2\angle 0°\text{A}$ for $I_2$ in the above equation.

  $\begin{array}{l}\left[16+j\left(9\times {10}^{-4}\right)I_1\right]+8\left(2\angle 0°\text{A}\right)=1\angle 0°\\ \left[16+j\left(9\times {10}^{-4}\right)I_1\right]=1\angle 0°-16\angle 0°\\ I_1=\frac{-15\angle 0°}{16+j\left(9\times {10}^{-4}\right)}\\ I_1=-0.9375+j\left(5.2734\times {10}^{-5}\right)\text{A}\end{array}$

Apply KVL to the loop on the left.

  $\begin{array}{l}-1\angle 0°+8I_1+V_{Th}=0\\ V_{Th}=-8I_1+1\angle 0°\end{array}$

Substitute $-0.9375+j\left(5.2734\times {10}^{-5}\right)\text{A}$ for $I_1$ in the above equation.

  $\begin{array}{c}{V}_{Th}=-8\left[-0.9375+j\left(5.2734\times {10}^{-5}\right)\right]+1\angle 0°\\ =8.5-j\left(42.872\times {10}^{-5}\right)\\ =\sqrt{{\left(8.5\right)}^2+{\left(42.872\times {10}^{-5}\right)}^2}{\tan }^{-1}\left(\frac{42.872\times {10}^{-5}}{8.5}\right)\\ =8.5\angle -0.003°\text{V}\end{array}$

To determine the Thevenin equivalent impedance, short circuit the voltage source and open circuit the current source and redraw the circuit.

The required diagram is shown in Figure 4

  

From the above figure, the expression for the Thevenin equivalent impedance is given by,

  $\begin{array}{c}{Z}_{Th}=\frac{\left(8\Omega \right)\left[8\Omega +j\left(9\times {10}^{-4}\right)\Omega \right]}{\left(8\Omega \right)+\left[8\Omega +j\left(9\times {10}^{-4}\right)\Omega \right]}\\ =\frac{64+j\left(72\times {10}^{-4}\right)}{16+j\left(9\times {10}^{-4}\right)}\\ =4+j\left(2.25\times {10}^4\right)\Omega \end{array}$

Mark the values and draw the Thevenin equivalent of the circuit.

The required diagram is shown in Figure 5

  

From the above figure the voltage across the capacitor is calculated as,

  $\begin{array}{c}{V}_C=\left(\frac{-j666.67\Omega }{-j666.67\Omega +j\left(2.25\times {10}^{-4}\right)\Omega }\right)\left(8.5\angle -0.003°\text{V}\right)\\ =\left(0.9999-j5.9997\times {10}^{-3}\right)\left(8.5\angle 0.003°\text{V}\right)\\ =\left(0.9999-j5.9997\times {10}^{-3}\right)\left(8.5\angle 0.003°\text{V}\right)\\ =8.5\angle 0.35°\text{V}\end{array}$ ....... (3)

The general form for the expression of voltage across the capacitor is given by,

  $v_C\left(t\right)=V_m\cos \left(\omega t+\varphi \right)$

From above and from equation (3), the general form of the voltage across the capacitor is given by,

  $v_C\left(t\right)=8.5\cos \left(\omega t-0.35°\right)$

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

  $v_C\left(t\right)=8.5\cos \left(300t-0.35°\right)$

Conclusion:

Therefore, the Thevenin equivalent of the network seen by the capacitor $C$ is shown in Figure 5. The value of the voltage $v_C\left(t\right)$ is $8.5\cos \left(300t-0.35°\right)$.