Chapter 13, Problem 57P

For the ideal transformer circuit of Fig. 13.122 below, find:

1. (a) I₁ and I₂,
2. (b) V₁, V₂, and V_o,
3. (c) the complex power supplied by the source.

To determine

Calculate the currents $I_1\text{and}{I}_2$ in the ideal transformer circuit in Figure 13.122.

Answer to Problem 57P

The value of currents $I_1\text{and}{I}_2$ are $\underset{\_}{25.9\angle 69.96°\text{A}\left(\text{rms}\right)}$ and $\underset{\_}{12.95\angle 69.96°\text{A}\left(\text{rms}\right)}$ respectively.

Explanation of Solution

Given data:

Refer to Figure 13.122 in the textbook for the transformer circuit.

The value of n from the given figure is 2.

Calculation:

From Figure 13.122, calculate the load impedance

$\begin{array}{c}{Z}_L=j3\parallel \left(12-j6\right)\\ =\frac{\left(j3\right)\left(12-j6\right)}{j3+\left(12-j6\right)}\\ =\frac{12+j54}{17}\Omega \end{array}$

In Figure 13.122, reflect the load of $\frac{12+j54}{17}\Omega$ to the primary side circuit. Consider the expression for the impedance ${Z^{\prime }}_L$.

${Z^{\prime }}_L=\frac{Z_L}{n^2}$

Substitute 2 for $n$ and $\frac{12+j54}{17}\Omega$ for $Z_L$.

$\begin{array}{c}{Z^{\prime }}_L=\frac{\frac{12+j54}{17}\Omega }{n^2}\\ =\frac{3+j13.5}{17}\Omega \end{array}$

Consider the expression for the impedance $Z_{\text{in}}$.

$Z_{\text{in}}=2\Omega +{Z^{\prime }}_L$

Substitute $\frac{3+j13.5}{17}\Omega$ for ${Z^{\prime }}_L$.

$\begin{array}{c}{Z}_{\text{in}}=2\Omega +\frac{3+j13.5}{17}\Omega \\ =2.3168\angle 20.04°\Omega \end{array}$

From Figure 1, write the expression for the current $I_1$.

$I_1=\frac{V}{Z_{\text{in}}}$

Substitute $2.3168\angle 20.04°\Omega$ for $Z_{\text{in}}$ and $60\angle 90°\text{V}$ for V.

$\begin{array}{c}{I}_1=\frac{60\angle 90°\text{V}}{2.3168\angle 20.04°\Omega }\\ =25.9\angle 69.96°\text{A}\left(\text{rms}\right)\end{array}$

Write the expression for the current $I_2$.

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

Substitute $25.9\angle 69.96°\text{A}$ for $I_1$ and 2 for n.

$\begin{array}{c}{I}_2=\frac{25.9\angle 69.96°\text{A}}{2}\\ =12.95\angle 69.96°\text{A}\left(\text{rms}\right)\end{array}$

Conclusion:

Thus, the value of currents $I_1\text{and}{I}_2$ are $\underset{\_}{25.9\angle 69.96°\text{A}\left(\text{rms}\right)}$ and $\underset{\_}{12.95\angle 69.96°\text{A}\left(\text{rms}\right)}$ respectively.

To determine

Calculate the voltages $V_1,V_2\text{and}{V}_o$.

Answer to Problem 57P

The value of voltages $V_1,V_2\text{and}{V}_o$ are $\underset{\_}{21.06\angle 147.44°\text{V}\left(\text{rms}\right)}$, $\underset{\_}{42.12\angle 147.44°\text{V}\left(\text{rms}\right)}$, and $\underset{\_}{42.12\angle 147.44°\text{V}\left(\text{rms}\right)}$ respectively.

Explanation of Solution

Calculation:

Apply Kirchhoff's voltage law to the primary loop contains current $I_1$ in Figure 1.

$60\angle 90°=2I_1+V_1$

Substitute $25.9\angle 69.96°\text{A}$ for $V_1$.

$\begin{array}{l}60\angle 90°=2\left(25.9\angle 69.96°\text{A}\right)+V_1\\ V_1=60\angle 90°-51.8\angle 69.96°\text{A}\\ V_1=j60-\left(17.7506+j48.6637\right)\\ V_1=-17.7506+j11.3363\end{array}$

$V_1=21.06\angle 147.44°\text{V}\left(\text{rms}\right)$

Write the expression for the voltage $V_1$.

$V_2=n{V}_1$

Substitute 2 for n and $21.06\angle 147.44°\text{V}$ for $V_1$.

$\begin{array}{c}{V}_2=2\left(21.06\angle 147.44°\text{V}\right)\\ =42.12\angle 147.44°\text{V}\left(\text{rms}\right)\end{array}$

Write the expression for the output voltage $V_1$ using Figure 13.122.

$V_o=V_2$

Substitute $42.12\angle 147.44°\text{V}\left(\text{rms}\right)$ for $V_2$.

$V_o=42.12\angle 147.44°\text{V}\left(\text{rms}\right)$

Conclusion:

Thus, The value of voltages $V_1,V_2\text{and}{V}_o$ are $\underset{\_}{21.06\angle 147.44°\text{V}\left(\text{rms}\right)}$, $\underset{\_}{42.12\angle 147.44°\text{V}\left(\text{rms}\right)}$, and $\underset{\_}{42.12\angle 147.44°\text{V}\left(\text{rms}\right)}$ respectively.

To determine

Calculate the complex power supplied by the source.

Answer to Problem 57P

The complex power supplied by the source is $\underset{\_}{1.554\angle 20.04°\text{kVA}}$.

Explanation of Solution

Calculation:

The conjugate of the current $I_1$ is,

$\begin{array}{c}{I}_1^{\ast }=25.9\angle -69.96°\text{A}\left(\text{rms}\right)\\ =8.8753-j24.3318\text{A}\left(\text{rms}\right)\end{array}$

Write the expression for the complex power supplied by the source.

$S=V{I}_1^{\ast }$

Write the Matlab code to find the required complex power.

V=i*60;

I1_C=8.8753-i*24.3318;

S=V*I1_C

The output of the Matlab code is given as follows.

S = 1459.91 + 532.52i

From the Matlab output, the complex power is,

$\begin{array}{c}S=1459.91+j532.52\text{VA}\\ =\text{1554}\angle \text{20}\text{.04}°\text{VA}\\ =1.554\angle \text{20}\text{.04}°\text{kVA}\end{array}$

Conclusion:

Thus, the complex power supplied by the source is $\underset{\_}{1.554\angle 20.04°\text{kVA}}$.