Chapter 13, Problem 24P

In the circuit of Fig. 13.93,

1. (a) find the coupling coefficient,
2. (b) calculate v_o,
3. (c) determine the energy stored in the coupled inductors at t = 2 s.

To determine

Calculate the coupling coefficient of the circuit in Figure 13.93.

Answer to Problem 24P

The coupling coefficient is $\underset{\_}{0.3535}$.

Explanation of Solution

Given data:

Refer to Figure 13.93 in the textbook for the circuit with coupled coils.

The value of $L_1,L_2,\text{and}M$ are 4 H, 2 H, and 1 H respectively.

Calculation:

Consider the expression for the coefficient of coupling in the coupled coils.

$k=\frac{M}{\sqrt{L_1{L}_2}}$

Substitute 1 H for M, 4 H for $L_1$, and 2 H for $L_2$.

$\begin{array}{c}k=\frac{1\text{H}}{\sqrt{\left(4\text{H}\right)\left(2\text{H}\right)}}\\ =\frac{1}{\sqrt{8}}\\ =0.3535\end{array}$

Conclusion:

Thus, the coupling coefficient is $\underset{\_}{0.3535}$.

To determine

Calculate the voltage $v_o$.

Answer to Problem 24P

The value of voltage $v_o$ is $\underset{\_}{321.6\cos \left(4t+57.6°\right)\text{mV}}$.

Explanation of Solution

Given data:

From Figure 13.93, the value of $\omega$ is $4\text{rad/s}$.

Calculation:

Write the expression for the inductive reactance.

$j{X}_L=j\omega L$        (1)

Write the expression for the capacitive reactance.

$-j{X}_C=\frac{1}{j\omega C}$        (2)

Substitute 4 H for $L$ and $4\text{rad/s}$ for $\omega$ in Equation (1).

$\begin{array}{c}j{X}_L=j\left(4\right)\left(4\text{rad/s}\right)\\ =j16\Omega \end{array}$

Substitute 2 H for $L$ and $4\text{rad/s}$ for $\omega$ in Equation (1).

$\begin{array}{c}j{X}_L=j\left(2\right)\left(4\text{rad/s}\right)\\ =j8\Omega \end{array}$

Substitute 1 H for $L$ and $4\text{rad/s}$ for $\omega$ in Equation (1).

$\begin{array}{c}j{X}_L=j\left(1\right)\left(4\text{rad/s}\right)\\ =j4\Omega \end{array}$

Substitute $\frac{1}{4}\text{F}$ for C and $4\text{rad/s}$ for $\omega$ in Equation (2).

$\begin{array}{c}-j{X}_C=\frac{1}{j\left(4\right)\left(\frac{1}{4}\text{F}\right)}\\ =-j\end{array}$

Calculate load impedance $Z_L$.

$\begin{array}{c}{Z}_L=1\Omega \parallel \left(-j\Omega \right)\\ =\frac{\left(1\Omega \right)\left(-j\Omega \right)}{1\Omega +\left(-j\Omega \right)}\\ =\frac{-j}{\left(1-j\right)\Omega }\\ =0.5\left(1-j\right)\Omega \end{array}$

Modify the Figure 13.93 by transforming the time-domain circuit with coupled-coils to frequency domain of the circuit with coupled-coils. The frequency domain equivalent circuit is shown in Figure 1.

From Figure 1, consider that the loops 1 and 2 contain the currents $I_1$ and $I_2$ respectively.

Apply Kirchhoff's voltage law to the loop 1 in Figure 1.

$\begin{array}{l}\left(2+j16\right)I_1+j4I_2=12\\ \left(1+j8\right)I_1+j2I_2=\frac{12}{2}\end{array}$

$\left(1+j8\right)I_1+j2I_2=6$        (3)

Apply Kirchhoff's voltage law to the loop 2 in Figure 1.

$\left(j8+0.5-j0.5\right)I_2+j4I_1=0$

$j4I_1+\left(j7.5+0.5\right)I_2=0$        (4)

Write equations (3) and (4) in matrix form as follows.

$\left[\begin{array}{cc}\left(1+j8\right)& j2\\ j4& \left(0.5+j7.5\right)\end{array}\right]\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{c}6\\ 0\end{array}\right]$        (5)

Write the MATLAB code to solve the equation (5).

A = [(1+j*8) j*2;j*4 (0.5+j*7.5)];

B = [6; 0];

I = inv(A)*B

The output in command window:

I =

   0.13036 - 0.84468i

  -0.09912 + 0.44389i

From the MATLAB output, the currents $I_1$ and $I_2$ are,

$\begin{array}{c}{I}_1=\left(0.13036-j0.84468\right)\text{A}\\ =0.855\angle -81.23°\text{A}\end{array}$

And

$\begin{array}{c}{I}_2=\left(-0.09912+0.44389\right)\text{A}\\ =0.455\angle 102.59°\text{A}\end{array}$

Write the expression for the voltage $V_o$.

$V_o=I_2\times 0.5\left(1-j\right)$

Substitute $0.455\angle 102.59°\text{A}$ for $I_2$.

$\begin{array}{c}{V}_o=\left(0.455\angle 102.59°\text{A}\right)\times \left(0.5-j0.5\right)\\ =\left(-0.09912+0.44389\right)\times \left(0.5-j0.5\right)\\ =0.1724+j0.2715\text{V}\\ =0.3217\angle 57.6°\mathrm{V}\end{array}$

Convert the phasor form to time domain form.

$\begin{array}{c}{v}_o=0.3216\cos \left(\omega t+57.6°\right)\text{V}\\ =321.6\cos \left(4t+57.6°\right)\text{mV}\left\{\because \omega =4\right\}\\ =321.6\cos \left(4t+57.6°\right)\text{mV}\end{array}$

Conclusion:

Thus, the value of voltage $v_o$ is $\underset{\_}{321.6\cos \left(4t+57.6°\right)\text{mV}}$.

To determine

Calculate the stored energy in the coupled coils at $t=2\text{ms}$.

Answer to Problem 24P

The energy stored in the coupled coils is $\underset{\_}{1.168\text{J}}$.

Explanation of Solution

Calculation:

From part (b), write the currents $I_1$ and $I_2$ in time-domain.

$i_1=0.855\cos \left(4t-81.23°\right)\text{A}$        (6)

$i_2=0.455\cos \left(4t+102.59°\right)\text{A}$        (7)

Substitute 2 s for t in Equation (6).

$\begin{array}{c}{i}_1=0.855\cos \left(4\left(2\right)-81.23°\right)\text{A}\\ =0.855\cos \left(\left(8\text{rad}\right)-81.23°\right)\text{A}\\ =0.855\cos \left(98.37°-81.23°\right)\text{A}\left\{\begin{array}{l}\because 8\text{rad}=8\times \frac{180}{\pi }\mathrm{deg}\\ =98.37°\end{array}\right\}\\ =0.817\text{A}\end{array}$

Substitute 2 s for t in Equation (7).

$\begin{array}{c}{i}_2=0.455\cos \left(4\left(2\right)+102.59°\right)\text{A}\\ =0.455\cos \left(\left(8\text{rad}\right)+102.59°\right)\text{A}\\ =0.455\cos \left(98.37°+102.59°\right)\text{A}\left\{\begin{array}{l}\because 8\text{rad}=8\times \frac{180}{\pi }\mathrm{deg}\\ =98.37°\end{array}\right\}\\ =-0.4249\text{A}\end{array}$

Write the expression for the total energy stored in the coupled coils.

$w=0.5L_1{i}_1^2+0.5L_2{i}_2^2+M{i}_1{i}_2$

Substitute 4 H for $L_1$, 2 H for $L_2$, 1 H for M, $0.817\text{A}$ for $i_1$, and $-0.4249\text{A}$ for $i_2$.

$\begin{array}{c}w=0.5\left(4\right){\left(0.817\text{A}\right)}^2+0.5\left(2\right){\left(-0.4249\text{A}\right)}^2+\left(1\right)\left(0.817\text{A}\right)\left(-0.4249\text{A}\right)\\ =1.334978+0.18054-0.3471433\\ =1.168\text{J}\end{array}$

Conclusion:

Thus, the energy stored in the coupled coils is $\underset{\_}{1.168\text{J}}$.