Chapter 9, Problem 41P

Find v(t) in the RLC circuit of Fig. 9.48.

Figure 9.48

To determine

Find the value the voltage $v\left(t\right)$ in the RLC circuit shown in Figure 9.48.

Answer to Problem 41P

The value of the voltage $v\left(t\right)$ in the RLC circuit shown in Figure 9.48 is $72.74\cos \left(t-18.43°\right)\text{V}$.

Explanation of Solution

Given date:

Refer to Figure 9.48 in the textbook.

Formula used:

Write the expression to convert the time domain expression into phasor domain.

$A\cos \left(\omega t+\varphi \right)\iff A\angle \varphi$        (1)

Here,

A is the magnitude,

$\omega$ is the angular frequency,

t is the time, and

$\varphi$ is the phase angle.

Write the expression to calculate the total phasor current.

$I=\frac{V_s}{Z_{\text{eq}}}$        (2)

Here,

$V_s$ is the phasor source voltage, and

$Z_{\text{eq}}$ is the total equivalent impedance.

Write the expression to calculate the impedance of the passive elements resistor, inductor and capacitor.

$Z_R=R$        (3)

$Z_L=j\omega L$        (4)

$Z_C=\frac{1}{j\omega C}$        (5)

Here,

$R$ is the value of the resistor,

$L$ is the value of the inductor, and

$C$ is the value of the capacitor.

Calculation:

The given RLC circuit is redrawn as Figure 1.

Refer to Figure 1, the source voltage is,

$v_s\left(t\right)=115\cos t\text{V}$

Here, angular frequency $\omega =1\frac{\text{rad}}{\text{s}}$.

Use the equation (1) to express the above equation in phasor form.

$V_s=\left(115\angle 0°\right)$

Substitute $1\Omega$ for $R_1$ in equation (3) to find $Z_{R_1}$.

$Z_{R_1}=1\Omega$

Substitute $1\Omega$ for $R_2$ in equation (3) to find $Z_{R_2}$.

$Z_{R_2}=1\Omega$

Substitute $1\text{H}$ for $L$ and $1\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (4) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(1\frac{\text{rad}}{\text{s}}\right)\left(1\text{H}\right)\\ =j\left(1\frac{\text{rad}}{\text{s}}\right)\left(1\Omega \cdot \text{s}\right)\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =j\Omega \end{array}$

Substitute $1\text{F}$ for $C$ and $1\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (5) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(1\frac{\text{rad}}{\text{s}}\right)\left(1\text{F}\right)}\\ =\frac{1}{j\left(1\right)\frac{\text{rad}}{\text{s}}\cdot \frac{\text{s}}{\Omega }}\left\{\because 1\text{F}=\frac{\text{1}\text{s}}{1\Omega }\right\}\\ =-j\Omega \end{array}$

The Figure 1 is redrawn as impedance circuit in the following Figure 2.

Refer to Figure 2, the series combination of the impedances $Z_{R_2}$ and $Z_L$ are connected in parallel with the impedance $Z_C$.

Write the expression to calculate the equivalent impedance 1 for the parallel combination as follows.

$Z_{\text{eq1}}=\left(Z_C\right)\parallel \left(Z_{R_2}+Z_L\right)$        (6)

Here,

$Z_{R_2}$ is the impedance of the resistor 2,

$Z_L$ is the impedance of the inductor, and

$Z_C$ is the impedance of the capacitor.

Substitute $1\Omega$ for $Z_{R_2}$, $j\Omega$ for $Z_L$ and $-j\Omega$ for $Z_C$ in equation (6) to find $Z_{\text{eq1}}$.

$\begin{array}{c}{Z}_{\text{eq1}}=\left(-j\Omega \right)\parallel \left(1\Omega +j\Omega \right)\\ =\left(-j\Omega \right)\parallel \left(1+j\right)\Omega \\ =\frac{\left(-j\Omega \right)\left(1\Omega +j\Omega \right)}{=-j\Omega +1\Omega +j\Omega }\\ =\left(1-j\right)\Omega \end{array}$

The reduced circuit of the Figure 2 is drawn as Figure 3.

Refer to Figure 2, the impedances $Z_{R_1}$ and $Z_{\text{eq1}}$ are connected in series form.

Write the expression to calculate the equivalent capacitance for the series connected impedances $Z_{R_1}$ and $Z_{\text{eq1}}$.

$Z_{\text{eq}}=Z_{R_1}+Z_{\text{eq1}}$        (7)

Here,

$Z_{R_1}$ is the impedance of the resistor 1, and

$Z_{\text{eq1}}$ is the equivalent impedance 1.

Substitute $1\Omega$ for $Z_{R_1}$, $\left(1-j\right)\Omega$ for $Z_{\text{eq1}}$ in equation (7) to find $Z_{\text{eq}}$.

$\begin{array}{c}{Z}_{\text{eq}}=1\Omega +\left(1-j\right)\Omega \\ =1\Omega +1\Omega -j\Omega \\ =2\Omega -j\Omega \\ =\left(2-j\right)\Omega \end{array}$

The reduced circuit of the Figure 3 is drawn as Figure 4.

Substitute $\left(115\angle 0°\right)\text{V}$ for $V_s$ and $\left(2-j\right)\Omega$ for $Z_{\text{eq}}$ in equation (2) to find $I$.

$\begin{array}{c}I=\frac{\left(115\angle 0°\right)\text{V}}{\left(2-j\right)\Omega }\\ =\frac{\left(115\angle 0°\right)\text{V}}{\left(2.236\angle -26.565°\right)\Omega }\\ =\left(51.431\angle 26.565°\right)\text{A}\end{array}$

Refer to Figure 2, the current through the capacitor is expressed as,

$I_C=\left(1+j\right)I$

Refer to Figure 2, the voltage across the capacitor is expressed as,

$V=Z_C{I}_C$

Substitute $-j\Omega$ for $Z_C$ and $\left(1+j\right)I$ for $I_C$ in above equation to find $V$.

$V=\left(-j\Omega \right)\left(\left(1+j\right)I\right)$

Substitute $\left(51.431\angle 26.565°\right)\text{A}$ for $I$ in above equation to find $V$.

$\begin{array}{c}V=\left(-j\Omega \right)\left(\left(1+j\right)\left(\left(51.431\angle 26.565°\right)\text{A}\right)\right)\\ =\left(1\angle -90°\right)\left(1.414\angle 45°\right)\left(51.431\angle 26.565°\right)\Omega \cdot \text{A}\\ =\left(\text{72}\text{.74}\angle -18.43°\right)\text{V}\end{array}$

Use the equation (1) to express the above equation in time domain form.

$v\left(t\right)=72.74\cos \left(\omega t-18.43°\right)\text{V}$

Substitute $1$ for $\omega$ in above equation to find $v\left(t\right)$.

$\begin{array}{c}v\left(t\right)=72.74\cos \left(\left(1\right)t-18.43°\right)\text{V}\\ =72.74\cos \left(t-18.43°\right)\text{V}\end{array}$

Conclusion:

Thus, the value of the voltage $v\left(t\right)$ in the RLC circuit shown in Figure 9.48 is $72.74\cos \left(t-18.43°\right)\text{V}$.