Chapter 13, Problem 54E

Obtain an expression for V₂/Vs in the circuit of Fig. 13.68 if (a) L₁ = 100 mH, L₂ = 500 mH, and M is its maximum possible value; (b) L₁ = 5L₂ = 1.4 H and k = 87% of its maximum possible value; (c) the two coils can be treated as an ideal transformer, the left-hand coil having 500 turns and the right-hand coil having 10,000 turns.

FIGURE 13.68

To determine

Find the expression for $\frac{V_2}{V_s}$ in the given circuit.

Answer to Problem 54E

The expression for $\frac{V_2}{V_s}$ is $\underset{\_}{\frac{-j4\sqrt{5}\omega }{200+j6.5\omega }}$.

Explanation of Solution

Given data:

Refer to Figure 13.68 in the textbook for the given circuit.

The circuit parameters are given as follows:

$\begin{array}{l}{L}_1=100\text{mH}\\ L_2=500\text{mH}\end{array}$

$M$ is the maximum possible value.

Formula used:

Write the expression for reactance due to inductive coil of self-inductance as follows:

$X_L=j\omega L$        (1)

Here,

$L$ is the self-inductance of inductive coil and

$\omega$ is the angular frequency of the inductive coil.

Write the expression for reactance due to inductive coil of mutual-inductance as follows:

$X_L=j\omega M$        (2)

Here,

$M$ is the mutual-inductance of the inductive coils

Write the expression for mutual inductance as follows:

$M=k\sqrt{L_1{L}_2}$        (3)

Here,

$k$ is the coefficient of coupling.

Calculation:

The maximum possible value of $M$ is possible only when the value of $k$ is 1.

Substitute 1 for $k$, $100\text{mH}$ for $L_1$, and $500\text{mH}$ for $L_2$ in Equation (3) to obtain the maximum value of $M$.

$\begin{array}{c}M=\left(1\right)\sqrt{\left(100\text{mH}\right)\left(500\text{mH}\right)}\\ =\sqrt{\left(0.1\right)\left(0.5\right)}\text{H}\\ =0.1\sqrt{5}\text{H}\end{array}$

Observer the dot notation in the circuit, use the expression in Equations (1), (2), and apply KVL to the primary winding-loop in the given circuit as follows:

$-V_s+\left(5\Omega \right)I_1+j\omega L_1{I}_1+j\omega M{I}_2=0$

Simplify the expression as follows:

$V_s=5I_1+j\omega L_1{I}_1+j\omega M{I}_2$

$V_s=\left(5+j\omega L_1\right)I_1+j\omega M{I}_2$        (4)

Substitute $100\text{mH}$ for $L_1$ and $0.1\sqrt{5}\text{H}$ for $M$ as follows:

$\begin{array}{c}{V}_s=\left[5+j\omega \left(100\text{mH}\right)\right]I_1+j\omega \left(0.1\sqrt{5}\text{H}\right)I_2\\ =\left[5+j\omega \left(0.1\right)\right]I_1+j0.1\sqrt{5}\omega I_2\end{array}$

$V_s=\left(5+j0.1\omega \right)I_1+j0.1\sqrt{5}\omega I_2$        (5)

Observer the dot notation, use the expression in Equations (1), (2), and apply KVL to the secondary winding-loop in the given circuit as follows:

$\begin{array}{l}j\omega L_2{I}_2+j\omega M{I}_1+\left(40\Omega \right)I_2=0\\ j\omega L_2{I}_2+j\omega M{I}_1+40I_2=0\\ \left(j\omega L_2+40\right)I_2+j\omega M{I}_1=0\end{array}$

$I_1=-\frac{\left(j\omega L_2+40\right)}{j\omega M}{I}_2$        (6)

Substitute $500\text{mH}$ for $L_2$ and $0.1\sqrt{5}\text{H}$ for $M$ as follows:

$\begin{array}{c}{I}_1=-\frac{\left[j\omega \left(500\text{mH}\right)+40\right]}{j\omega \left(0.1\sqrt{5}\text{H}\right)}{I}_2\\ =-\frac{\left(40+j0.5\omega \right)}{j0.1\sqrt{5}\omega }{I}_2\end{array}$

Substitute $\left[-\frac{\left(40+j0.5\omega \right)}{j0.1\sqrt{5}\omega }{I}_2\right]$ for $I_1$ in Equation (5) as follows:

$\begin{array}{c}{V}_s=\left(5+j0.1\omega \right)\left[-\frac{\left(40+j0.5\omega \right)}{j0.1\sqrt{5}\omega }{I}_2\right]+j0.1\sqrt{5}\omega I_2\\ =\left[-\frac{\left(5+j0.1\omega \right)\left(40+j0.5\omega \right)}{j0.1\sqrt{5}\omega }+j0.1\sqrt{5}\omega \right]I_2\\ =\left(\frac{200+j2.5\omega +j4\omega -0.05{\omega }^2+0.05{\omega }^2}{-j0.1\sqrt{5}\omega }\right)I_2\end{array}$

$V_s=\left(\frac{200+j6.5\omega }{-j0.1\sqrt{5}\omega }\right)I_2$        (7)

From the given circuit, write the expression for $V_2$ as follows:

$V_2=40I_2$

Rearrange the expression as follows:

$I_2=\frac{V_2}{40}$

Substitute $\frac{V_2}{40}$ for $I_2$ in Equation (7) as follows:

$\begin{array}{c}{V}_s=\left(\frac{200+j6.5\omega }{-j0.1\sqrt{5}\omega }\right)\left(\frac{V_2}{40}\right)\\ =\left(\frac{200+j6.5\omega }{-j4\sqrt{5}\omega }\right)V_2\end{array}$

Rearrange the expression as follows:

$\frac{V_2}{V_s}=\frac{-j4\sqrt{5}\omega }{200+j6.5\omega }$

Conclusion:

Thus, the expression for $\frac{V_2}{V_s}$ is $\underset{\_}{\frac{-j4\sqrt{5}\omega }{200+j6.5\omega }}$.

To determine

Find the expression for $\frac{V_2}{V_s}$ in the given circuit.

Answer to Problem 54E

The expression for $\frac{V_2}{V_s}$ is $\underset{\_}{\frac{-j21.789\omega }{200+j57.4\omega -0.0955{\omega }^2}}$.

Explanation of Solution

Given data:

The circuit parameters are given as follows:

$\begin{array}{l}{L}_1=5L_2\\ L_1=1.4\text{H}\\ k=0.87\end{array}$

Calculation:

Find the value of $L_2$ as follows:

$\begin{array}{c}{L}_2=\frac{L_1}{5}\\ =\frac{1.4\text{H}}{5}\\ =0.28\text{H}\end{array}$

Substitute 0.87 for $k$, 1.4 H for $L_1$, and 0.28 H for $L_2$ in Equation (3) to obtain the maximum value of $M$.

$\begin{array}{c}M=\left(0.87\right)\sqrt{\left(1.4\text{H}\right)\left(0.28\text{H}\right)}\\ =\left(0.87\right)\left(0.626\right)\text{H}\\ =0.5446\text{H}\end{array}$

Substitute 1.4 H for $L_1$ and 0.5446 H for $M$ in Equation (5) as follows:

$V_s=\left[5+j\omega \left(1.4\text{H}\right)\right]I_1+j\omega \left(0.5446\text{H}\right)I_2$

$V_s=\left(5+j1.4\omega \right)I_1+j0.5446\omega I_2$        (8)

Substitute 0.28 H for $L_2$ and 0.5446 H for $M$ in Equation (6) as follows:

$\begin{array}{c}{I}_1=-\frac{\left[j\omega \left(\text{0}\text{.28}\text{H}\right)+40\right]}{j\omega \left(0.5446\text{H}\right)}{I}_2\\ =-\frac{\left(40+j0.28\omega \right)}{j0.5446\omega }{I}_2\end{array}$

Substitute $\left[-\frac{\left(40+j0.28\omega \right)}{j0.5446\omega }{I}_2\right]$ for $I_1$ in Equation (8) as follows:

$\begin{array}{c}{V}_s=\left(5+j1.4\omega \right)\left[-\frac{\left(40+j0.28\omega \right)}{j0.5446\omega }{I}_2\right]+j0.5446\omega I_2\\ =\left[-\frac{\left(5+j1.4\omega \right)\left(40+j0.28\omega \right)}{j0.5446\omega }+j0.5446\omega \right]I_2\\ =\left(\frac{200+j1.4\omega +j56\omega -0.392{\omega }^2+0.2965{\omega }^2}{-j0.5446\omega }\right)I_2\\ =\left(\frac{200+j57.4\omega -0.0955{\omega }^2}{-j0.5446\omega }\right)I_2\end{array}$

Substitute $\frac{V_2}{40}$ for $I_2$ as follows:

$\begin{array}{c}{V}_s=\left(\frac{200+j57.4\omega -0.0955{\omega }^2}{-j0.5446\omega }\right)\left(\frac{V_2}{40}\right)\\ =\left(\frac{200+j57.4\omega -0.0955{\omega }^2}{-j21.789\omega }\right)V_2\end{array}$

Rearrange the expression as follows:

$\frac{V_2}{V_s}=\frac{-j21.789\omega }{200+j57.4\omega -0.0955{\omega }^2}$

Conclusion:

Thus, the expression for $\frac{V_2}{V_s}$ is $\underset{\_}{\frac{-j21.789\omega }{200+j57.4\omega -0.0955{\omega }^2}}$.

To determine

Find the expression for $\frac{V_2}{V_s}$ in the given circuit.

Answer to Problem 54E

The expression for $\frac{V_2}{V_s}$ is $\underset{\_}{\frac{20}{51}}$.

Explanation of Solution

Given data:

The circuit parameters are given as follows:

$\begin{array}{l}{N}_2=10000\\ N_1=500\end{array}$

Formula used:

Write the expression for transformer ratio as follows:

$a=\frac{N_2}{N_1}$        (9)

Here,

$N_1$ is the number of turns of primary winding of the transformer and

$N_2$ is the number of turns of secondary winding of the transformer.

Write the expression for input impedance of the transformer as follows:

$Z_{\text{in}}=\frac{Z_L}{a^2}$        (10)

Here,

$Z_L$ is the load impedance, which is $40\Omega$ and

$a$ is the transformer ratio.

From the given circuit, write the expression for current through primary winding of the transformer as follows:

$I_1=\frac{V_s}{5\Omega +Z_{\text{in}}}$        (11)

Calculation:

Substitute 500 for $N_1$ and 10000 for $N_2$ in Equation (9) as follows:

$\begin{array}{c}a=\frac{10000}{500}\\ =20\end{array}$

Substitute 20 for $a$ and $40\Omega$ for $Z_L$ in Equation (10) to obtain the value of input impedance of the circuit.

$\begin{array}{c}{Z}_{\text{in}}=\frac{40\Omega }{{\left(20\right)}^2}\\ =0.1\Omega \end{array}$

Substitute $0.1\Omega$ for $Z_{\text{in}}$ in Equation (11) as follows:

$I_1=\frac{V_s}{5\Omega +0.1\Omega }$

$I_1=\frac{V_s}{5.1\Omega }$        (12)

From the given circuit, write the expression for $V_1$ as follows:

$V_1=V_s-I_1\left(5\Omega \right)$

From Equation (12), substitute $\frac{V_s}{5.1\Omega }$ for $I_1$ as follows:

$\begin{array}{c}{V}_1=V_s-\left(\frac{V_s}{5.1\Omega }\right)\left(5\Omega \right)\\ =\left(\frac{5.1\Omega -5\Omega }{5.1\Omega }\right)V_s\end{array}$

$V_1=\left(\frac{1}{51}\right)V_s$        (13)

Write the expression for transformer ratio in terms of voltages as follows:

$a=-\frac{V_2}{V_1}$

Substitute 20 for $a$ and form Equation (13), substitute $\left(\frac{1}{51}\right)V_s$ for $V_1$ as follows:

$20=-\frac{V_2}{\left(\frac{1}{51}\right)V_s}$

Rearrange the expression as follows:

$\frac{V_2}{V_s}=-\frac{20}{51}$

Conclusion:

Thus, the expression for $\frac{V_2}{V_s}$ is $-\underset{\_}{\frac{20}{51}}$.