Chapter 13, Problem 15E

In the circuit of Fig. 13.43, M is reduced by an order of magnitude. Calculate i₃ if v₁ = 10 cos (800t − 20°) V.

FIGURE 13.43

To determine

Find the value of $i_3\left(t\right)$.

Answer to Problem 15E

The value of $i_3\left(t\right)$ is $\underset{\_}{1.88\angle 52.62°\text{A}}$.

Explanation of Solution

Given data:

Refer to Figure 13.43 in the textbook for the given circuit.

$v_1\left(t\right)=8\sin \left(720t-20°\right)\text{V}$

The value of mutual inductance is reduced by an order of magnitude. Therefore, write the value of mutual inductance as follows:

$M=50\text{nH}$

Formula used:

Write the expression for reactance due to 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 capacitance as follows:

$X_C=\frac{-j}{\omega C}$        (2)

Here,

$C$ is the capacitance of the capacitor.

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

$X_M=j\omega M$        (3)

Here,

$M$ is the mutual-inductance of the inductive coil.

Calculation:

From the given source voltage, write the value of angular frequency and the source voltage $V_1$ as follows:

$\begin{array}{l}{v}_1\left(t\right)=8\sin \left(720t-20°\right)\text{V}\\ \omega =800\text{rad}/\text{s}\\ V_1=8\angle -20°\text{V}\end{array}$

Rewrite the expression for $V_1$ as follows:

$V_1=\left(7.5175-j2.7361\right)\text{V}$

Substitute $1\text{mH}$ for $L$ and $800\text{rad}/\text{s}$ for $\omega$ in Equation (1) to obtain the reactance due to $1\text{mH}$ inductor as follows:

$\begin{array}{c}{X}_{1\text{mH}}=j\left(800\text{rad}/\text{s}\right)\left(1\text{mH}\right)\\ =j\left(800\right)\left(1\times {10}^{-3}\right)\Omega \\ =j0.8\end{array}$

Substitute $2\text{mH}$ for $L$ and $800\text{rad}/\text{s}$ for $\omega$ in Equation (1) to obtain the reactance due to $2\text{mH}$ inductor as follows:

$\begin{array}{c}{X}_{2\text{mH}}=j\left(800\text{rad}/\text{s}\right)\left(2\text{mH}\right)\\ =j\left(800\right)\left(2\times {10}^{-3}\right)\Omega \\ =j1.6\end{array}$

Substitute $1\text{mF}$ for $C$ and $800\text{rad}/\text{s}$ for $\omega$ in Equation (2) to obtain the reactance due to $1\text{mF}$ capacitor as follows:

$\begin{array}{c}{X}_{1\text{mF}}=\frac{-j}{\left(800\text{rad}/\text{s}\right)\left(1\text{mF}\right)}\\ =\frac{-j}{\left(800\right)\left(1\times {10}^{-3}\right)}\Omega \\ =-j1.25\Omega \end{array}$

Substitute $50\text{nH}$ for $M$ and $800\text{rad}/\text{s}$ for $\omega$ in Equation (3) to obtain the reactance due to $500\text{nH}$ mutual inductance of the inductive coils as follows:

$\begin{array}{c}{X}_{50\text{nH}}=j\left(800\text{rad}/\text{s}\right)\left(50\text{nH}\right)\\ =j\left(800\right)\left(50\times {10}^{-9}\right)\Omega \\ =j\left(40\times {10}^{-6}\right)\Omega \end{array}$

Use the obtained values of reactance due to inductor and capacitors and draw the phasor representation for the given circuit as shown in Figure 1.

Apply KVL to the loop-$I_1$ in the circuit in Figure 1 as follows:

$-V_1+\left(1.8\Omega \right)I_1-{V^{\prime }}_1=0$

Neglect the dimensions and write the expression as follows:

$-V_1+1.8I_1-{V^{\prime }}_1=0$        (4)

Apply KVL to the loop-$I_2$ in the given circuit as follows:

$+{V^{\prime }}_1-j1.25I_2+{V^{\prime }}_2=0$        (5)

Apply KVL to the loop-$I_3$ in the given circuit as follows:

$-{V^{\prime }}_2+\left(2\Omega \right)I_3=0$

Neglect the dimensions and write the expression as follows:

$-{V^{\prime }}_2+2I_3=0$        (6)

From the circuit in Figure 1, write the expression for ${V^{\prime }}_1$ as follows:

${V^{\prime }}_1=j0.8\left(I_2-I_1\right)+j\left(40\times {10}^{-6}\right)\left(I_2-I_3\right)$

${V^{\prime }}_1=-j0.8I_1+j0.80004I_2-j\left(40\times {10}^{-6}\right)I_3$        (7)

From the circuit in Figure 1, write the expression for ${V^{\prime }}_2$ as follows:

${V^{\prime }}_2=j1.6\left(I_2-I_3\right)+j\left(40\times {10}^{-6}\right)\left(I_2-I_1\right)$

${V^{\prime }}_2=-j\left(40\times {10}^{-6}\right)I_1+j1.60004I_2-j1.6I_3$        (8)

Rewrite the expression in Equation (4) as follows:

$1.8I_1=V_1+{V^{\prime }}_1$

Substitute $\left(7.5175-j2.7361\right)\text{V}$ for $V_1$ and $\left(-j0.8I_1+j0.80004I_2-j\left(40\times {10}^{-6}\right)I_3\right)$ for ${V^{\prime }}_1$ as follows:

$1.8I_1=\left[\begin{array}{l}\left(7.5175-j2.7361\right)\text{V}+\\ \left(-j0.8I_1+j0.80004I_2-j\left(40\times {10}^{-6}\right)I_3\right)\end{array}\right]$

Simplify the expression as follows:

$I_2=\frac{\left(1.8+j0.8\right)I_1+j\left(40\times {10}^{-6}\right)I_3-\left(7.5175-j2.7361\right)}{j0.80004}$        (9)

Rearrange the expression in Equation (6) as follows:

$2I_3={V^{\prime }}_2$

Substitute $\left[-j\left(40\times {10}^{-6}\right)I_1+j1.60004I_2-j1.6I_3\right]$ for ${V^{\prime }}_2$ as follows:

$2I_3=-j\left(40\times {10}^{-6}\right)I_1+j1.60004I_2-j1.6I_3$

Simplify the expression as follows:

$I_2=\frac{j\left(40\times {10}^{-6}\right)I_1+\left(2+j1.6\right)I_3}{j1.60004}$        (10)

From Equation (9), substitute $\frac{\left(1.8+j0.8\right)I_1+j\left(40\times {10}^{-6}\right)I_3-\left(7.5175-j2.7361\right)}{j0.80004}$ for $I_2$ as follows:

$\frac{\left(1.8+j0.8\right)I_1+j\left(40\times {10}^{-6}\right)I_3-\left(7.5175-j2.7361\right)}{j0.80004}=\frac{j\left(40\times {10}^{-6}\right)I_1+\left(2+j1.6\right)I_3}{j1.60004}$

Simplify the expression as follows:

$I_1=\frac{\left(1+j0.8\right)I_3+\left(7.5175-j2.7361\right)}{1.8+j0.8}$        (11)

Substitute $\frac{\left(1+j0.8\right)I_3+\left(7.5175-j2.7361\right)}{1.8+j0.8}$ for $I_1$ in Equation (10) as follows:

$I_2=\frac{j\left(40\times {10}^{-6}\right)\left[\frac{\left(1+j0.8\right)I_3+\left(7.5175-j2.7361\right)}{1.8+j0.8}\right]+\left(2+j1.6\right)I_3}{j1.60004}$

Simplify the expression as follows:

$I_2\cong \left(2+j1.6\right)I_3+\left(0.00007-j0.00007\right)$        (12)

Rearrange the expression in Equation (5) as follows:

$j1.25I_2={V^{\prime }}_1+{V^{\prime }}_2$

Substitute $\left[-j0.8I_1+j0.80004I_2-j\left(40\times {10}^{-6}\right)I_3\right]$ for ${V^{\prime }}_1$ and $\left[-j\left(40\times {10}^{-6}\right)I_1+j1.60004I_2-j1.6I_3\right]$ for ${V^{\prime }}_2$ as follows:

$\begin{array}{l}j1.25I_2=\left[\begin{array}{l}-j0.8I_1+j0.80004I_2-j\left(40\times {10}^{-6}\right)I_3+\\ \left[-j\left(40\times {10}^{-6}\right)I_1+j1.60004I_2-j1.6I_3\right]\end{array}\right]\\ j1.25I_2=-j0.80004I_1+j2.40008I_2-j1.60004I_3\\ j1.15008I_2=j0.80004I_1+j1.60004I_3\end{array}$

From Equation (11), substitute $\frac{\left(1+j0.8\right)I_3+\left(7.5175-j2.7361\right)}{1.8+j0.8}$ for $I_1$ as follows:

$\begin{array}{c}j1.15008I_2=j0.80004\left[\frac{\left(1+j0.8\right)I_3+\left(7.5175-j2.7361\right)}{1.8+j0.8}\right]+j1.60004I_3\\ =\left(-0.1319+j2.1031\right)I_3+\left(2.2555+j2.3388\right)\end{array}$

Simplify the expression as follows:

$\begin{array}{c}{I}_2=\frac{\left(-0.1319+j2.1031\right)I_3+\left(2.2555+j2.3388\right)}{j1.15008}\\ =\left(1.8286+j0.1146\right)I_3+\left(2.0335-j1.9611\right)\end{array}$

Substitute $\left(1.8286+j0.1146\right)I_3+\left(2.0335-j1.9611\right)$ for $I_2$ in Equation (12) and simplify the expression as follows:

$\begin{array}{l}\left(1.8286+j0.1146\right)I_3+\left(2.0335-j1.9611\right)=\left(2+j1.6\right)I_3+\left(0.00007-j0.00007\right)\\ \left(0.1714+j1.4854\right)I_3=2.0334-j1.9610\end{array}$

Simplify the expression as follows;

$\begin{array}{c}{I}_3=\frac{2.0334-j1.9610}{0.1714+j1.4854}\\ =-1.1469-j1.5012\end{array}$

Rewrite the expression in polar form to obtain the value of $i_3\left(t\right)$.

$i_3\left(t\right)=1.88\angle 52.62°\text{A}$

Conclusion:

Thus, the value of $i_3\left(t\right)$ is $\underset{\_}{1.88\angle 52.62°\text{A}}$.