Chapter 13, Problem 44E

(a)

To determine

Find the voltages $v_1$ and $v_2$ in the given circuit.

Answer to Problem 44E

The values of voltages $v_1$ and $v_2$ in the given circuit are $\underset{\_}{-0.687\text{V}}$ and $\underset{\_}{-1.585\text{V}}$, respectively.

Explanation of Solution

Given data:

Refer to Figure 13.64 in the textbook for the given circuit.

Formula used:

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

$a=\frac{V_2}{V_1}$        (1)

Here,

$V_1$ is the voltage across primary winding of the transformer and

$V_2$ is the voltage across secondary winding of the transformer.

Write the expression for equivalent resistance of secondary winding when the transformer is referred to the primary winding as follows:

$R_2{}^{\prime }=\frac{R_2}{a^2}$        (2)

Here,

$R_2$ is the resistance of the secondary winding of the transformer.

Calculation:

Consider left side transformer as first transformer and right side transformer in the given circuit as second transformer to obtain the required objectives.

From the given source current, write the maximum value of current $I_s$ as follows:

$\begin{array}{l}{i}_s=25\cos \left(120\pi t\right)\text{mA}\\ I_s=25\text{mA}\end{array}$

From the given circuit, the transformer ratio for two transformers are written as follows:

$\begin{array}{c}{a}_1=\frac{1}{5}\\ a_2=\frac{15}{2}\end{array}$

Use the current division rule, the expression in Equation (2), and write the expression for current through $2\Omega$ resistor in the given circuit as follows:

$I_{2\Omega }=\frac{2.7\text{k}\Omega }{2.7\text{k}\Omega +2\Omega +\frac{\left(4\Omega +\frac{100\Omega }{{\left(a_2\right)}^2}\right)}{{\left(a_1\right)}^2}}{I}_s$

Substitute $25\text{mA}$ for $I_s$, $\frac{1}{5}$ for $a_1$, and $\frac{15}{2}$ for $a_2$ to obtain the value of current through $2\Omega$ resistor.

$\begin{array}{c}{I}_{2\Omega }=\frac{2.7\text{k}\Omega }{2.7\text{k}\Omega +2\Omega +\frac{\left(4\Omega +\frac{100\Omega }{{\left(\frac{15}{2}\right)}^2}\right)}{{\left(\frac{1}{5}\right)}^2}}\left(25\text{mA}\right)\\ =\left(\frac{2700\Omega }{2702\Omega +144.4444\Omega }\right)\left(25\text{mA}\right)\\ =23.7137\text{mA}\\ \cong 23.71\text{mA}\end{array}$

From the given circuit, write the expression for current through $2.7\text{k}\Omega$ resistor as follows:

$I_{2.7\text{k}\Omega }=I_s-I_{2\Omega }$

Substitute $25\text{mA}$ for $I_s$ and $23.71\text{mA}$ for $I_{2\Omega }$ to obtain the value of current through $2.7\text{k}\Omega$ resistor.

$\begin{array}{c}{I}_{2.7\text{k}\Omega }=25\text{mA}-23.71\text{mA}\\ =1.29\text{mA}\end{array}$

From the given circuit, write the expression for voltage across primary winding of the first transformer as follows:

${\left(V_1\right)}_1=\left(I_{2.7\text{k}\Omega }\right)\left(2.7\text{k}\Omega \right)-\left(I_{2\Omega }\right)\left(2\Omega \right)$

Substitute $1.29\text{mA}$ for $I_{2.7\text{k}\Omega }$ and $23.71\text{mA}$ for $I_{2\Omega }$ to obtain the value of voltage across primary winding of the first transformer.

$\begin{array}{c}{\left(V_1\right)}_1=\left(1.29\text{mA}\right)\left(2.7\text{k}\Omega \right)-\left(23.71\text{mA}\right)\left(2\Omega \right)\\ =\left(1.29\times {10}^{-3}\text{A}\right)\left(2.7\times {10}^3\Omega \right)-\left(23.71\times {10}^{-3}\text{A}\right)\left(2\Omega \right)\\ =3.483\text{V}-0.04742\text{V}\\ =3.43558\text{V}\end{array}$

Use the expression in Equation (1) and write the expression for voltage across secondary winding of the first transformer as follows:

${\left(V_2\right)}_1=a_1{\left(V_1\right)}_1$

Substitute $\frac{1}{5}$ for $a_1$ and 3.43558 V for ${\left(V_1\right)}_1$ to obtain the value of ${\left(V_2\right)}_1$ in the given circuit.

$\begin{array}{c}{\left(V_2\right)}_1=\left(\frac{1}{5}\right)\left(3.43558\text{V}\right)\\ =0.687\text{V}\end{array}$

From the given circuit observe the dot convention and write the expression for $v_1$ as follows:

$v_1=-{\left(V_2\right)}_1$

Substitute 0.687 V for ${\left(V_2\right)}_1$ to obtain the value of $v_1$ as follows:

$v_1=-0.687\text{V}$

Use voltage division rule, expression in Equation (2), and write the expression for voltage across primary winding of the second transformer as follows:

${\left(V_1\right)}_2=v_1\left[\frac{\frac{100\Omega }{{\left(a_2\right)}^2}}{4\Omega +\frac{100\Omega }{{\left(a_2\right)}^2}}\right]$

Substitute $\frac{15}{2}$ for $a_2$ and $\left(-0.687\text{V}\right)$ for $v_1$ to obtain the value of voltage across primary winding of the second transformer.

$\begin{array}{c}{\left(V_1\right)}_2=\left(-0.687\text{V}\right)\left[\frac{\frac{100\Omega }{{\left(\frac{15}{2}\right)}^2}}{4\Omega +\frac{100\Omega }{{\left(\frac{15}{2}\right)}^2}}\right]\\ =\left(-0.687\text{V}\right)\left(\frac{1.7777\Omega }{5.7777\Omega }\right)\\ =-0.2113\text{V}\end{array}$

From the given circuit, write the expression for $v_2$ as follows:

$v_2=a_2{\left(V_1\right)}_2$

Substitute $\frac{15}{2}$ for $a_2$ and $\left(-0.2113\text{V}\right)$ for ${\left(V_1\right)}_2$ to obtain the value of $v_2$ in the given circuit.

$\begin{array}{c}{v}_2=\left(\frac{15}{2}\right)\left(-0.2113\text{V}\right)\\ \cong -1.585\text{V}\end{array}$

Conclusion:

Thus, the values of voltages $v_1$ and $v_2$ in the given circuit are $\underset{\_}{-0.687\text{V}}$ and $\underset{\_}{-1.585\text{V}}$, respectively.

(b)

To determine

Find the average power delivered to each resistor in the given circuit.

Answer to Problem 44E

The average power delivered to $2.7\text{k}\Omega$, $2\Omega$, $4\Omega$, and $100\Omega$ resistors are $\underset{\_}{2.25\text{mW},0.562\text{mW},28.3\text{mW}}$, and $\underset{\_}{12.56\text{mW}}$, respectively.

Explanation of Solution

Formula used:

Write the expression for average power delivered to resistor as follows:

$P=\frac{1}{2}\frac{{\left(V_m\right)}^2}{R}$        (3)

Here,

$V_m$ is the maximum value of the voltage across the resistor and

$R$ is the resistance of the resistor.

Calculation:

From Part (a), the voltage across $2.7\text{k}\Omega$ resistor is calculated as follows:

$V_{2.7\text{k}\Omega }=I_{2.7\text{k}\Omega }\left(2.7\text{k}\Omega \right)$

Substitute $1.29\text{mA}$ for $I_{2.7\text{k}\Omega }$ as follows:

$\begin{array}{c}{V}_{2.7\text{k}\Omega }=\left(1.29\text{mA}\right)\left(2.7\text{k}\Omega \right)\\ =\left(1.29\times {10}^{-3}\right)\left(2.7\times {10}^3\right)\\ =3.483\text{V}\end{array}$

Write the expression for voltage across $2\Omega$ resistor as follows:

$V_{2\Omega }=I_{2\Omega }\left(2\Omega \right)$

Substitute $23.71\text{mA}$ for $I_{2\Omega }$ as follows:

$\begin{array}{c}{V}_{2\Omega }=\left(23.71\text{mA}\right)\left(2\Omega \right)\\ =\left(23.71\times {\text{10}}^{-3}\right)\left(2\right)\text{V}\\ =0.04742\text{V}\end{array}$

Write the expression for voltage across $4\Omega$ resistor as follows:

$V_{4\Omega }=v_1-{\left(V_1\right)}_2$

From Part (a), substitute $\left(-0.687\text{V}\right)$ for $v_1$ and $\left(-0.2113\text{V}\right)$ for ${\left(V_1\right)}_2$ as follows:

$\begin{array}{c}{V}_{4\Omega }=\left(-0.687\text{V}\right)-\left(-0.2113\text{V}\right)\\ =-0.4757\text{V}\end{array}$

Write the expression for voltage across $100\Omega$ resistor as follows:

$V_{100\Omega }=v_2$

From Part (a), substitute $\left(-1.585\text{V}\right)$ for $v_2$ as follows:

$V_{100\Omega }=-1.585\text{V}$

Modify the expression in Equation (3) to obtain the average power delivered to $2.7\text{k}\Omega$ resistor as follows:

$P_{2.7\text{k}\Omega }=\frac{1}{2}\frac{{\left(V_{2.7\text{k}\Omega }\right)}^2}{2.7\text{k}\Omega }$

Substitute 3.483 V for $V_{2.7\text{k}\Omega }$ to obtain the average power delivered to $2.7\text{k}\Omega$ resistor.

$\begin{array}{c}{P}_{2.7\text{k}\Omega }=\frac{1}{2}\frac{{\left(3.483\text{V}\right)}^2}{2.7\text{k}\Omega }.\\ =\left(\frac{1}{2}\right)\frac{{\left(3.483\right)}^2}{2700}\text{W}\\ \cong 2.25\times {10}^{-3}\text{W}\\ \cong 2.25\text{mW}\end{array}$

Modify the expression in Equation (3) to obtain the average power delivered to $2\Omega$ resistor as follows:

$P_{2\Omega }=\frac{1}{2}\frac{{\left(V_{2\Omega }\right)}^2}{2\Omega }$

Substitute 0.04742 V for $V_{2\Omega }$ to obtain the average power delivered to $2\Omega$ resistor.

$\begin{array}{c}{P}_{2\Omega }=\frac{1}{2}\frac{{\left(0.15\text{V}\right)}^2}{2\Omega }\\ \cong 0.562\times {10}^{-3}\text{W}\\ \cong 0.562\text{mW}\end{array}$

Modify the expression in Equation (3) to obtain the average power delivered to $4\Omega$ resistor as follows:

$P_{4\Omega }=\frac{1}{2}\frac{{\left(V_{4\Omega }\right)}^2}{4\Omega }$

Substitute $-0.4757\text{V}$ for $V_{4\Omega }$ to obtain the average power delivered to $4\Omega$ resistor.

$\begin{array}{c}{P}_{4\Omega }=\frac{1}{2}\frac{{\left(-0.4757\text{V}\right)}^2}{4\Omega }\\ \cong 28.3\times {10}^{-3}\text{W}\\ \cong 28.3\text{mW}\end{array}$

Modify the expression in Equation (3) to obtain the average power delivered to $100\Omega$ resistor as follows:

$P_{100\Omega }=\frac{1}{2}\frac{{\left(V_{100\Omega }\right)}^2}{100\Omega }$

Substitute $\left(-1.585\text{V}\right)$ for $V_{100\Omega }$ to obtain the average power delivered to $100\Omega$ resistor.

$\begin{array}{c}{P}_{100\Omega }=\frac{1}{2}\frac{{\left(-1.585\text{V}\right)}^2}{100\Omega }\\ \cong 12.56\times {10}^{-3}\text{W}\\ \cong 12.56\text{mW}\end{array}$

Conclusion:

Thus, the average power delivered to $2.7\text{k}\Omega$, $2\Omega$, $4\Omega$, and $100\Omega$ resistors are $\underset{\_}{2.25\text{mW},0.562\text{mW},28.3\text{mW}}$, and $\underset{\_}{12.56\text{mW}}$, respectively.