Chapter 13, Problem 1E

(a)

To determine

Find the steady-state expression for $v_1$ when $i_1=0\text{A}$ and $i_2=5\cos \left(8t\right)\text{A}$.

Answer to Problem 1E

The steady-state expression for $v_1$ is $\underset{\_}{-0.04\sin \left(8t\right)\text{V}}$.

Explanation of Solution

Given data:

Refer to Figure 13.35 in the textbook for the given circuit.

$\begin{array}{l}{L}_1=10\text{mH}\\ L_2=5\text{mH}\\ M=1\text{mH}\end{array}$

$\begin{array}{l}{i}_1=0\text{A}\\ i_2=5\cos \left(8t\right)\text{A}\end{array}$

Formula used:

Write the expression for steady-state expression for $v_1$ in the given circuit as follows:

$v_1=L_1\frac{d{i}_1}{dt}+M\frac{d{i}_2}{dt}$        (1)

Here,

$L_1$ is the self-inductance of primary coil and

$M$ is the mutual inductance of the primary and secondary coils.

Calculation:

Substitute $10\text{mH}$ for $L_1$, $1\text{mH}$ for $M$, 0 A for $i_1$, and $5\cos \left(8t\right)\text{A}$ for $i_2$ in Equation (1) to obtain the steady-state expression for $v_1$.

$\begin{array}{c}{v}_1=\left(10\text{mH}\right)\left[\frac{d}{dt}\left(0\text{A}\right)\right]+\left(1\text{mH}\right)\left[\frac{d}{dt}\left(5\cos \left(8t\right)\text{A}\right)\right]\\ =0\text{V}+\left[\left(1\times {10}^{-3}\right)\left(5\right)\left(-8\right)\sin \left(8t\right)\right]\text{V}\\ =-0.04\sin \left(8t\right)\text{V}\end{array}$

Conclusion:

Thus, the steady-state expression for $v_1$ is $\underset{\_}{-0.04\sin \left(8t\right)\text{V}}$.

(b)

To determine

Find the steady-state expression for $v_2$ when $i_1=3\sin \left(100t\right)\text{A}$ and $i_2=0\text{A}$.

Answer to Problem 1E

The steady-state expression for $v_2$ is $\underset{\_}{0.3\cos \left(100t\right)\text{V}}$.

Explanation of Solution

Given data:

$\begin{array}{l}{L}_2=5\text{mH}\\ i_1=3\sin \left(100t\right)\text{A}\\ i_2=0\text{A}\end{array}$

Formula used:

Write the expression for steady-state expression for $v_2$ in the given circuit as follows:

$v_2=L_2\frac{d{i}_2}{dt}+M\frac{d{i}_1}{dt}$        (2)

Here,

$L_2$ is the self-inductance of secondary coil.

Calculation:

Substitute $5\text{mH}$ for $L_2$, $1\text{mH}$ for $M$, $3\sin \left(100t\right)\text{A}$ for $i_1$, and 0 A for $i_2$ in Equation (2) to obtain the steady-state expression for $v_2$.

$\begin{array}{c}{v}_2=\left(5\text{mH}\right)\left[\frac{d}{dt}\left(0\text{A}\right)\right]+\left(1\text{mH}\right)\left[\frac{d}{dt}\left(3\sin \left(100t\right)\text{A}\right)\right]\\ =0\text{V}+\left[\left(1\times {10}^{-3}\right)\left(3\right)\left(100\right)\cos \left(100t\right)\right]\text{V}\\ =0.3\cos \left(100t\right)\text{V}\end{array}$

Conclusion:

Thus, the steady-state expression for $v_2$ is $\underset{\_}{0.3\cos \left(100t\right)\text{V}}$.

(c)

To determine

Find the steady-state expression for $v_2$ when $i_1=5\cos \left(8t-40°\right)\text{A}$ and $i_2=4\sin \left(8t\right)\text{A}$.

Answer to Problem 1E

The steady-state expression for $v_2$ is $\underset{\_}{0.16\cos \left(8t\right)-0.04\sin \left(8t-40°\right)\text{V}}$.

Explanation of Solution

Given data:

$\begin{array}{l}{i}_1=5\cos \left(8t-40°\right)\text{A}\\ i_2=4\sin \left(8t\right)\text{A}\end{array}$

Calculation:

Substitute $5\text{mH}$ for $L_2$, $1\text{mH}$ for $M$, $5\cos \left(8t-40°\right)\text{A}$ for $i_1$, and $4\sin \left(8t\right)\text{A}$ for $i_2$ in Equation (2) to obtain the steady-state expression for $v_2$.

$\begin{array}{c}{v}_2=\left(5\text{mH}\right)\left[\frac{d}{dt}\left(4\sin \left(8t\right)\text{A}\right)\right]+\left(1\text{mH}\right)\left[\frac{d}{dt}\left(5\cos \left(8t-40°\right)\text{A}\right)\right]\\ =\left[\left(5\times {10}^{-3}\right)\left(4\right)\left(8\right)\cos \left(8t\right)\right]\text{V}+\left[\left(1\times {10}^{-3}\right)\left(5\right)\left(-8\right)\sin \left(8t-40°\right)\right]\text{V}\\ =0.16\cos \left(8t\right)-0.04\sin \left(8t-40°\right)\text{V}\end{array}$

Conclusion:

Thus, the steady-state expression for $v_2$ is $\underset{\_}{0.16\cos \left(8t\right)-0.04\sin \left(8t-40°\right)\text{V}}$.