Chapter 9, Problem 36P

Using Fig. 9.43, design a problem to help other students better understand impedance.

Figure 9.43

To determine

Design a problem to make better understand about the impedance using Figure 9.43.

Explanation of Solution

Problem design:

Determine the value of current $i$ in the circuit in Figure 9.43. Assume that voltage is,

$v_s=60\cos \left(200t-10°\right)\text{V}$.

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 phasor current.

$I=\frac{V}{Z}$        (2)

Here,

$V$ is the phasor voltage, and

$Z$ is the total 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 Figure 9.43 is redrawn as Figure 1 by assuming the values for the passive elements.

 

Given voltage equation is,

$v_s=60\cos \left(200t-10°\right)$

Here, angular frequency $\omega =200\frac{\text{rad}}{\text{s}}$.

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

$V=\left(60\angle -10°\right)$

Substitute $2\text{k}\Omega$ for $R_1$ in equation (3) to find $Z_{R_1}$.

$Z_{R_1}=2\text{k}\Omega$

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

$Z_{R_2}=1\text{k}\Omega$

Substitute $1\text{k}\Omega$ for $R_3$ in equation (3) to find $Z_{R_3}$.

$Z_{R_3}=1\text{k}\Omega$

Substitute $100\text{mH}$ for $L$ and $200\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (4) to find $Z_L$.

$\begin{array}{c}{Z}_L=j\left(200\frac{\text{rad}}{\text{s}}\right)\left(100\text{mH}\right)\\ =j\left(200\frac{\text{rad}}{\text{s}}\right)\left(100\times {10}^{-3}\text{H}\right)\left\{\because 1\text{m}={10}^{-3}\right\}\\ =j\left(200\frac{\text{rad}}{\text{s}}\right)\left(0.1\Omega \cdot \text{s}\right)\left\{\because 1\text{H}=1\Omega \cdot 1\text{s}\right\}\\ =j20\Omega \end{array}$

Substitute $10\mu \text{F}$ for $C$ and $200\frac{\text{rad}}{\text{s}}$ for $\omega$ in equation (5) to find $Z_C$.

$\begin{array}{c}{Z}_C=\frac{1}{j\left(200\frac{\text{rad}}{\text{s}}\right)\left(10\mu \text{F}\right)}\\ =\frac{1}{j\left(200\right)\left(10\times {10}^{-6}\right)\frac{\text{rad}}{\text{s}}\cdot \text{F}}\left\{\because 1\mu ={10}^{-6}\right\}\\ =\frac{1}{j\left(2\times {10}^{-3}\right)\frac{\text{rad}}{\text{s}}\cdot \frac{\text{s}}{\Omega }}\left\{\because 1\text{F}=\frac{\text{1}\text{s}}{1\Omega }\right\}\\ =-j500\Omega \end{array}$

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

 

Refer to Figure 2, the impedances $Z_C$ and $Z_{R_3}$ are connected in parallel form.

Write the expression to calculate the equivalent capacitance 1 for the parallel connected impedances $Z_C$ and $Z_{R_3}$.

$Z_{\text{eq1}}=\frac{Z_C{Z}_{R_3}}{Z_C+Z_{R_3}}$        (6)

Here,

$Z_C$ is the impedance of the capacitor, and

$Z_{R_3}$ is the impedance of the resistor 3.

Substitute $-j500\Omega$ for $Z_C$ and $1\text{k}\Omega$ for $Z_{R_3}$ in equation (6) to find $Z_{\text{eq1}}$.

$\begin{array}{c}{Z}_{\text{eq1}}=\frac{\left(-j500\Omega \right)\left(1\text{k}\Omega \right)}{-j500\Omega +1\text{k}\Omega }\\ =\frac{\left(-j500\Omega \right)\left(1\times {10}^3\Omega \right)}{\left(-j500\Omega \right)+\left(1\times {10}^3\Omega \right)}\left\{\because 1\text{k}={10}^3\right\}\\ =\frac{\left(1000\Omega \right)\left(-j500\Omega \right)}{\left(1000-j500\right)\Omega }\\ =\left(200-j400\right)\Omega \end{array}$

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

 

Refer to Figure 3, the impedances $Z_L$ and $Z_{\text{eq1}}$ are connected in series form.

Write the expression to calculate the equivalent capacitance 2 for the series connected impedances $Z_L$ and $Z_{\text{eq1}}$.

$Z_{\text{eq2}}=Z_L+Z_{\text{eq1}}$        (7)

Here,

$Z_L$ is the impedance of the inductor, and

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

Substitute $j20\Omega$ for $Z_L$ and $\left(200-j400\right)\Omega$ for $Z_{\text{eq1}}$ in equation (7) to find $Z_{\text{eq2}}$.

$\begin{array}{c}{Z}_{\text{eq2}}=j20\Omega +\left(200-j400\right)\Omega \\ =j20\Omega +200\Omega -j400\Omega \\ =200\Omega -j380\Omega \\ =\left(200-j380\right)\Omega \end{array}$

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

 

Refer to Figure 4, the impedances $Z_{R_2}$ and $Z_{\text{eq2}}$ are connected in parallel form.

Write the expression to calculate the equivalent capacitance 3 for the parallel connected impedances $Z_{R_2}$ and $Z_{\text{eq2}}$.

$Z_{\text{eq3}}=\frac{Z_{R_2}{Z}_{\text{eq2}}}{Z_{R_2}+Z_{\text{eq2}}}$        (8)

Here,

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

$Z_{\text{eq2}}$ is the equivalent impedance 2.

Substitute $1\text{k}\Omega$ for $Z_{R_2}$ and $\left(200-j380\right)\Omega$ for $Z_{\text{eq2}}$ in equation (8) to find $Z_{\text{eq3}}$.

$\begin{array}{c}{Z}_{\text{eq3}}=\frac{\left(1\text{k}\Omega \right)\left(\left(200-j380\right)\Omega \right)}{\left(1\text{k}\Omega \right)+\left(\left(200-j380\right)\Omega \right)}\\ =\frac{\left(1\times {10}^3\Omega \right)\left(200-j380\right)\Omega }{\left(1\times {10}^3\Omega \right)+\left(\left(200-j380\right)\Omega \right)}\left\{\because 1\text{k}={10}^3\right\}\\ =\frac{\left(1000\Omega \right)\left(200-j380\right)\Omega }{1200\Omega -j380\Omega }\\ =\left(242.62-j239.84\right)\Omega \end{array}$

The reduced circuit of the Figure 4 is drawn as Figure 5.

 

Refer to Figure 5, the impedances $Z_{R_1}$ and $Z_{\text{eq3}}$ are connected in series form.

Write the expression to calculate the equivalent capacitance 4 for the series connected impedances $Z_{R_1}$ and $Z_{\text{eq3}}$.

$Z_{\text{eq4}}=Z_{R_1}+Z_{\text{eq3}}$        (9)

Here,

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

$Z_{\text{eq3}}$ is the equivalent impedance 3.

Substitute $2\text{k}\Omega$ for $Z_{R_1}$ and $\left(242.62-j239.84\right)\Omega$ for $Z_{\text{eq3}}$ in equation (9) to find $Z_{\text{eq4}}$.

$\begin{array}{c}{Z}_{\text{eq4}}=2\text{k}\Omega +\left(242.62-j239.84\right)\Omega \\ =\left(2\times {10}^3\Omega \right)+\left(242.62-j239.84\right)\Omega \left\{\because 1\text{k}={10}^3\right\}\\ =2000\Omega +242.62\Omega -j239.84\Omega \\ =\left(2242.62-j239.84\right)\Omega \end{array}$

The reduced circuit of the Figure 5 is drawn as Figure 6.

 

Therefore, the equivalent impedance of the circuit in Figure 1 is,

$\begin{array}{c}Z=\left(2242.62-j239.84\right)\Omega \\ =\left(2255\angle -6.104°\right)\Omega \end{array}$

Substitute $\left(60\angle -10°\right)$ for $V$ and $\left(2255\angle -6.104°\right)$ for $Z$ in equation (2) to find $I$.

$\begin{array}{c}I=\frac{\left(60\angle -10°\right)}{\left(2255\angle -6.104°\right)}\\ =\left(0.02661\angle -3.896°\right)\text{A}\\ =\left(0.02661\times {10}^3\times {10}^{-3}\angle -3.896°\right)\text{A}\\ =\left(26.61\angle -3.896°\right)\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

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

$i=26.61\cos \left(\omega t-3.896°\right)\text{mA}$

Substitute $200$ for $\omega$ in above equation to find $i\left(t\right)$.

$\begin{array}{c}i=26.61\cos \left(\left(200\right)t-3.896°\right)\text{mA}\\ =26.61\cos \left(200t-3.896°\right)\text{mA}\end{array}$

Therefore, the value of current $\left(i\right)$ in the circuit in Figure 9.43 is $26.61\cos \left(200t-3.896°\right)\text{mA}$.

Conclusion:

Thus, the problem to make better understand about the impedance using Figure 9.43 is designed.