Chapter 13, Problem 46P

1. (a) Find I₁ and I₂ in the circuit of Fig. 13.111 below.
2. (b) Switch the dot on one of the windings. Find I₁ and I₂ again.

To determine

Calculate the value of currents $I_1\text{and}{I}_2$ in the given circuit.

Answer to Problem 46P

The currents $I_1\text{and}{I}_2$ are $\underset{\_}{1.072\angle 5.868°\text{A}}$ and $\underset{\_}{0.536\angle 185.868°\text{A}}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 13.111 in the textbook for the transformer circuit with given values.

Calculation:

In Figure 13.111, reflect the secondary circuit to the primary circuit. Consider the expression to find the input impedance $Z_{\text{in}}$.

$Z_{\text{in}}=\left(R_1+j\omega L_1\right)\Omega +\frac{Z_L}{{\left(n\right)}^2}$

Write the Matlab code to find the input impedance.

n=2;

w=1;

R1=10;

L1=16;

ZL=12-i*8;

Z1=R1+i*(w*L1);

Zin= Z1+(ZL/n^2)

The Matlab output is given below.

Zin = 13 + 14i

From the Matlab output, the input impedance is,

$Z_{\text{in}}=13+j14\Omega$

Calculate the secondary side voltage source after reflection.

$V=\frac{10\angle 30°\text{V}}{-n}$

Substitute $2$ for $n$.

$\begin{array}{c}V=\frac{10\angle 30°\text{V}}{-2}\\ =-5\angle 30°\text{V}\end{array}$

The reduced circuit from Figure 13.111is drawn and it is shown in Figure 1.

From Figure 1, write the expression using Kirchhoff’s voltage law.

$-16\angle 60°+Z_{\text{in}}{I}_{\text{1}}-5\angle 30°=0$

Substitute $13+j14\Omega$ for $Z_{\text{in}}$.

$\begin{array}{l}-16\angle 60°+\left(13+j14\Omega \right)I_{\text{1}}-5\angle 30°=0\\ I_{\text{1}}=\frac{\left(16\angle 60°+5\angle 30°\right)}{\left(13+j14\Omega \right)}\\ I_{\text{1}}=\frac{\left(8+j13.8564+4.3301+j2.5\right)}{\left(13+j14\Omega \right)}\\ I_{\text{1}}=\frac{12.3301+j16.3564}{13+j14\Omega }\end{array}$

$\begin{array}{c}{I}_{\text{1}}=\frac{20.4832\angle 52.9896°}{19.1050\angle 47.1211°}\\ =1.072\angle 5.868°\text{A}\end{array}$

Write the expression for the current $I_2$.

$I_2=-\frac{I_1}{n}$

Substitute $1.072\angle 5.868°\text{A}$ for $I_1$ and 2 for n.

$\begin{array}{c}{I}_2=-\frac{1.072\angle 5.868°\text{A}}{2}\\ =-0.536\angle 5.868°\text{A}\\ =0.536\angle 185.868°\text{A}\end{array}$

Conclusion:

Thus, the currents $I_1\text{and}{I}_2$ are $\underset{\_}{1.072\angle 5.868°\text{A}}$ and $\underset{\_}{0.536\angle 185.868°\text{A}}$ respectively.

To determine

Calculate the value of currents $I_1\text{and}{I}_2$ in the given circuit when the dot is switched on one of the windings.

Answer to Problem 46P

The currents $I_1\text{and}{I}_2$ are $\underset{\_}{0.625\angle 25°\text{A}}$ and $\underset{\_}{0.3125\angle 25°\text{A}}$ respectively.

Explanation of Solution

Calculation:

Consider that the dot in the secondary side is switched to the down side. Switching a dot will not affect the impedance $Z_{\text{in}}$ but it will affect the currents $I_{\text{1}}$ and $I_{\text{2}}$.

Calculate the secondary side voltage source after reflection when the dot is switched.

$V=\frac{10\angle 30°\text{V}}{n}$

Substitute $2$ for $n$.

$\begin{array}{c}V=\frac{10\angle 30°\text{V}}{2}\\ =5\angle 30°\text{V}\end{array}$

The reduced circuit from Figure 13.111 is drawn and it is shown in Figure 2.

From Figure 1, write the expression using Kirchhoff’s voltage law.

$-16\angle 60°+Z_{\text{in}}{I}_{\text{1}}+5\angle 30°=0$

Substitute $13+j14\Omega$ for $Z_{\text{in}}$.

$\begin{array}{l}-16\angle 60°+\left(13+j14\Omega \right)I_{\text{1}}+5\angle 30°=0\\ I_{\text{1}}=\frac{\left(16\angle 60°-5\angle 30°\right)}{\left(13+j14\Omega \right)}\\ I_{\text{1}}=\frac{\left(8+j13.8564-\left(4.3301+j2.5\right)\right)}{\left(13+j14\Omega \right)}\\ I_{\text{1}}=\frac{3.6699+j11.3564}{13+j14\Omega }\end{array}$

$\begin{array}{c}{I}_{\text{1}}=\frac{11.9347\angle 72.0914°}{19.1050\angle 47.1211°}\\ =0.625\angle 25°\text{A}\end{array}$

Write the expression for the current $I_2$ when the secondary side dot is switched.

$I_2=\frac{I_1}{n}$

Substitute $0.625\angle 25°\text{A}$ for $I_1$ and 2 for n.

$\begin{array}{c}{I}_2=\frac{0.625\angle 25°\text{A}}{2}\\ =0.3125\angle 25°\text{A}\end{array}$

Conclusion:

Thus, the currents $I_1\text{and}{I}_2$ are $\underset{\_}{0.625\angle 25°\text{A}}$ and $\underset{\_}{0.3125\angle 25°\text{A}}$ respectively.