Chapter 13, Problem 27E

Consider the circuit represented in Fig. 13.53. The coupling coefficient k = 0.75. If i_s = 5 cos 200t mA, calculate the total energy stored at t = 0 and t = 5 ms if (a) a-b is open-circuited (as shown); (b) a-b is short-circuited.

FIGURE 13.53

To determine

Find the total energy stored in the system at $t=0\text{ms}$ and $t=5\text{ms}$ when a-b terminals are open circuited in the circuit in Figure 13.53 in the textbook.

Answer to Problem 27E

The total energy stored in the system at $t=0\text{ms}$ and $t=5\text{ms}$ when a-b terminals are open circuited in the circuit are $\underset{\_}{52.5\text{nJ}}$ and $\underset{\_}{15.31\text{nJ}}$, respectively.

Explanation of Solution

Given data:

Refer to Figure 13.53 in the textbook for the given circuit.

$\begin{array}{l}k=0.75\\ i_s=5\cos \left(200t\right)\text{mA}\end{array}$

The terminals a-b in the given circuit are open circuited.

Formula used:

Write the expression for energy stored in the magnetic field due to self-inductance of the coil at an instant of time as follows:

$w\left(t\right)=\frac{1}{2}L{\left[i\left(t\right)\right]}^2$        (1)

Here,

$i\left(t\right)$ is the current through an inductive coil and

$L$ is the inductance of the coil.

Calculation:

As the terminals a-b in the given circuit are open circuited, the current through the secondary winding loop is 0 A. Therefore, the total energy stored in the system is only due to the primary coils of the given circuit.

From the given circuit, find the value of inductance of primary coil when the terminals a-b in the given circuit are open circuited as follows:

$\begin{array}{c}L=3\text{mH}+1.2\text{mH}\\ =4.2\text{mH}\end{array}$

Find the current through the primary coil at $t=0\text{ms}$ as follows:

$\begin{array}{l}{i}_s\left(t\right)=5\cos \left(200t\right)\text{mA}\\ i_s\left(0\text{ms}\right)=5\cos \left[\left(200\right)\left(0\text{ms}\right)\right]\text{mA}\end{array}$

$\begin{array}{c}{i}_s\left(0\text{ms}\right)=5\cos \left[\left(200\right)\left(0\times {10}^{-3}\right)\right]\text{mA}\\ =5\left(1\right)\text{mA}\\ =5\text{mA}\end{array}$

Modify the expression in Equation (1) for energy stored in the system due to the coil $L_1$ at $t=0\text{ms}$ as follows:

$w_{\text{total}}\left(0\text{ms}\right)=\frac{1}{2}L{\left[i_s\left(0\text{ms}\right)\right]}^2$

Substitute $4.2\text{mH}$ for $L$ and $5\text{mA}$ for $i_s\left(0\text{ms}\right)$ to obtain the energy stored in the system at $t=0\text{ms}$ as follows:

$\begin{array}{c}{w}_{\text{total}}\left(0\text{ms}\right)=\left(\frac{1}{2}\right)\left(4.2\text{mH}\right){\left(5\text{mA}\right)}^2\\ =\left(\frac{1}{2}\right)\left(4.2\times {10}^{-3}\right)\left(25\times {10}^{-6}\right)\text{J}\\ =52.5\times {10}^{-9}\text{J}\\ =52.5\text{nJ}\end{array}$

Find the current through the primary coil at $t=5\text{ms}$ as follows:

$\begin{array}{l}{i}_s\left(t\right)=5\cos \left(200t\right)\text{mA}\\ i_s\left(5\text{ms}\right)=5\cos \left[\left(200\right)\left(5\text{ms}\right)\right]\text{mA}\end{array}$

$\begin{array}{c}{i}_s\left(5\text{ms}\right)=5\cos \left[\left(200\right)\left(5\times {10}^{-3}\right)\right]\text{mA}\\ =5\left(0.5403\right)\text{mA}\\ \cong 2.7\text{mA}\end{array}$

Modify the expression in Equation (1) for energy stored in the system due to the coil $L_1$ at $t=5\text{ms}$ as follows:

$w_{\text{total}}\left(5\text{ms}\right)=\frac{1}{2}{L}_1{\left[i_s\left(5\text{ms}\right)\right]}^2$

Substitute $4.2\text{mH}$ for $L_1$ and $2.7\text{mA}$ for $i_s\left(5\text{ms}\right)$ to obtain the energy stored in the system at $t=5\text{ms}$ as follows:

$\begin{array}{c}{w}_{\text{total}}\left(5\text{ms}\right)=\left(\frac{1}{2}\right)\left(4.2\text{mH}\right){\left(2.7\text{mA}\right)}^2\\ =\left(\frac{1}{2}\right)\left(4.2\times {10}^{-3}\right)\left(7.29\times {10}^{-6}\right)\text{J}\\ \cong 15.31\times {10}^{-9}\text{J}\\ \cong 15.31\text{nJ}\end{array}$

Conclusion:

Thus, the total energy stored in the system at $t=0\text{ms}$ and $t=5\text{ms}$ when a-b terminals are open circuited in the circuit are $\underset{\_}{52.5\text{nJ}}$ and $\underset{\_}{15.31\text{nJ}}$, respectively.

To determine

Find the total energy stored in the system at $t=0\text{ms}$ and $t=5\text{ms}$ when a-b terminals are short-circuited in the circuit in Figure 13.53 in the textbook.

Answer to Problem 27E

The total energy stored in the system at $t=0\text{ms}$ and $t=5\text{ms}$ when a-b terminals are short-circuited in the circuit are $\underset{\_}{44.1\text{nJ}}$ and $\underset{\_}{12.8\text{nJ}}$, respectively.

Explanation of Solution

Given data:

The terminals a-b in the given circuit are short-circuited.

$\begin{array}{l}{L}_1=1.2\text{mH}\\ L_1=12\text{mH}\\ k=0.75\end{array}$

Formula used:

Write the expression for energy stored in the magnetic field due to mutual inductance of the coils at an instant of time as follows:

$w\left(t\right)=M{i}_s\left(t\right)i_2\left(t\right)$        (2)

Here,

$i_2\left(t\right)$ is the current through the secondary coil at an instant of time $t$.

Write the expression for mutual inductance in terms of self-inductance of primary and secondary coils as follows:

$M=k\sqrt{L_1{L}_2}$        (3)

Here,

$k$ is the coupling coefficient,

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

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

Calculation:

Substitute $1.2\text{mH}$ for $L_1$, $1.2\text{mH}$ for $L_2$, and 0.75 for $k$ in Equation (3) to obtain the value of mutual inductance in the given circuit.

$\begin{array}{c}M=\left(0.75\right)\sqrt{\left(1.2\text{mH}\right)\left(12\text{mH}\right)}\\ =2.8460\text{mH}\\ \cong 2.85\text{mH}\end{array}$

From the given circuit, find the current $i_2\left(t\right)$ when the terminals a-b in the given circuit are short-circuited as follows:

$i_2\left(t\right)=\frac{-j\omega M}{R_2+j\omega L_2}{i}_s\left(t\right)$

Substitute $100\text{m}\Omega$ for $R_2$, $200\text{rad}/\text{s}$ for $\omega$, $2.85\text{mH}$ for $M$, $1.2\text{mH}$ for $L_2$, and $5\cos \left(200t\right)\text{mA}$ for $i_s\left(t\right)$ as follows:

$\begin{array}{c}{i}_2\left(t\right)=\left[\frac{-j\left(200\text{rad}/\text{s}\right)\left(2.85\text{mH}\right)}{100\text{m}\Omega +j\left(200\text{rad}/\text{s}\right)\left(12\text{mH}\right)}\right]\left[5\cos \left(200t\right)\text{mA}\right]\\ =\left[\frac{-j\left(200\right)\left(0.00285\right)}{0.1+j\left(200\right)\left(0.012\right)}\right]\left[5\cos \left(200t\right)\text{mA}\right]\\ =\left[\frac{-j0.57}{0.1+j2.4}\right]\left[5\cos \left(200t\right)\text{mA}\right]\end{array}$

Simplify the expression as follows:

$\begin{array}{c}{i}_2\left(t\right)=\left(-0.2370-j0.0098\right)\left[5\cos \left(200t\right)\text{mA}\right]\\ =\left(0.2372\angle -177.6°\right)\left[5\cos \left(200t\right)\text{mA}\right]\\ =1.186\cos \left(200t-177.6°\right)\text{mA}\end{array}$

Find the current through the secondary coil at $t=0\text{ms}$ as follows:

$\begin{array}{l}{i}_2\left(t\right)=1.186\cos \left(200t-177.6°\right)\text{mA}\\ i_2\left(0\text{ms}\right)=1.186\cos \left(200\left(0\text{ms}\right)-177.6°\right)\text{mA}\end{array}$

$\begin{array}{c}{i}_2\left(0\text{ms}\right)=1.186\left(-9991\right)\text{mA}\\ \cong -1.186\text{mA}\end{array}$

From Part (a), the energy stored in the system due to the coil $L_1$ at $t=0\text{ms}$ is $52.5\text{nJ}$.

$w_1\left(0\text{ms}\right)=52.5\text{nJ}$

Modify the expression in Equation (1) for energy stored in the system due to the coil $L_2$ at $t=0\text{ms}$ as follows:

$w_2\left(0\text{ms}\right)=\frac{1}{2}{L}_2{\left[i_2\left(0\text{ms}\right)\right]}^2$

Substitute $12\text{mH}$ for $L_2$ and $\left(-1.186\text{mA}\right)$ for $i_2\left(0\text{ms}\right)$ to obtain the energy stored in the system due to the coil $L_2$ at $t=0\text{ms}$ as follows:

$\begin{array}{c}{w}_2\left(0\text{ms}\right)=\left(\frac{1}{2}\right)\left(12\text{mH}\right){\left(-1.186\text{mA}\right)}^2\\ =\left(\frac{1}{2}\right)\left(12\times {10}^{-3}\right)\left(1.4065\times {10}^{-6}\right)\text{J}\\ =8.439\times {10}^{-9}\text{J}\\ =8.439\text{nJ}\end{array}$

Rearrange the expression in Equation (2) to find the energy stored in the magnetic field due to the mutual inductance of the coils at $t=0\text{ms}$ as follows:

$w_M\left(0\text{ms}\right)=M{i}_s\left(0\text{ms}\right)i_2\left(0\text{ms}\right)$

Substitute $2.85\text{mH}$ for $M$, $5\text{mA}$ for $i_s\left(0\text{ms}\right)$, and $\left(-1.186\text{mA}\right)$ for $i_2\left(0\text{ms}\right)$ to obtain the energy stored in the magnetic field due to mutual inductance of the coils at $t=0\text{ms}$.

$\begin{array}{c}{w}_M\left(0\text{ms}\right)=\left(2.85\text{mH}\right)\left(5\text{mA}\right)\left(-1.186\text{mA}\right)\\ =-16.9005\text{nJ}\end{array}$

Write the expression for total energy stored in the system at $t=0\text{ms}$ as follows:

$w\left(0\text{ms}\right)=w_1\left(0\text{ms}\right)+w_2\left(0\text{ms}\right)+w_M\left(0\text{ms}\right)$        (4)

Substitute $52.5\text{nJ}$ for $w_1\left(0\text{ms}\right)$, $8.439\text{nJ}$ for $w_2\left(0\text{ms}\right)$, and $-16.9005\text{nJ}$ J for $w_M\left(0\text{ms}\right)$ to obtain the total energy stored in the system at $t=0\text{ms}$.

$\begin{array}{c}w\left(0\text{ms}\right)=52.5\text{nJ}+8.439\text{J}+\left(-16.9005\right)\\ =44.0385\text{nJ}\\ \cong 44.1\text{nJ}\end{array}$

From Part (a), the energy stored in the system due to the coil $L_1$ at $t=5\text{ms}$ is $15.31\text{nJ}$.

$w_1\left(5\text{ms}\right)=15.31\text{nJ}$

Find the current through the secondary coil at $t=5\text{ms}$ as follows:

$\begin{array}{l}{i}_2\left(t\right)=1.186\cos \left(200t-177.6°\right)\text{mA}\\ i_2\left(5\text{ms}\right)=1.186\cos \left(200\left(5\text{ms}\right)-177.6°\right)\text{mA}\end{array}$

$\begin{array}{c}{i}_2\left(5\text{ms}\right)=1.186\cos \left[200\left(0.005\right)-\frac{177.6°\left(\pi \right)}{180°}\right]\text{mA}\\ =1.186\cos \left(1-3.0997\right)\text{mA}\\ \cong -0.5984\text{mA}\end{array}$

Modify the expression in Equation (1) for energy stored in the system due to the coil $L_2$ at $t=5\text{ms}$ as follows:

$w_2\left(5\text{ms}\right)=\frac{1}{2}{L}_2{\left[i_2\left(5\text{ms}\right)\right]}^2$

Substitute $12\text{mH}$ for $L_2$ and $\left(-0.5984\text{mA}\right)$ for $i_2\left(0\text{ms}\right)$ to obtain the energy stored in the system due to the coil $L_2$ at $t=5\text{ms}$ as follows:

$\begin{array}{c}{w}_2\left(5\text{ms}\right)=\left(\frac{1}{2}\right)\left(12\text{mH}\right){\left(-0.5984\text{mA}\right)}^2\\ =\left(\frac{1}{2}\right)\left(12\times {10}^{-3}\right)\left(0.3580\times {10}^{-6}\right)\text{J}\\ =2.148\times {10}^{-9}\text{J}\\ =2.148\text{nJ}\end{array}$

Rearrange the expression in Equation (2) to find the energy stored in the magnetic field due to the mutual inductance of the coils at $t=5\text{ms}$ as follows:

$w_M\left(5\text{ms}\right)=M{i}_s\left(5\text{ms}\right)i_2\left(5\text{ms}\right)$

Substitute $2.85\text{mH}$ for $M$, $2.7\text{mA}$ for $i_s\left(5\text{ms}\right)$, and $\left(-0.5984\text{mA}\right)$ for $i_2\left(5\text{ms}\right)$ to obtain the energy stored in the magnetic field due to mutual inductance of the coils at $t=5\text{ms}$.

$\begin{array}{c}{w}_M\left(5\text{ms}\right)=\left(2.85\text{mH}\right)\left(2.7\text{mA}\right)\left(-0.5984\text{mA}\right)\\ =-4.6046\text{nJ}\end{array}$

Modify the expression in Equation (4) for total energy stored in the system at $t=5\text{ms}$ as follows:

$w\left(5\text{ms}\right)=w_1\left(5\text{ms}\right)+w_2\left(5\text{ms}\right)+w_M\left(5\text{ms}\right)$

Substitute $15.31\text{nJ}$ for $w_1\left(5\text{ms}\right)$, $2.148\text{nJ}$ for $w_2\left(5\text{ms}\right)$, and $-4.6046\text{nJ}$ J for $w_M\left(5\text{ms}\right)$ to obtain the total energy stored in the system at $t=5\text{ms}$.

$\begin{array}{c}w\left(5\text{ms}\right)=15.31\text{nJ}+2.148\text{nJ}+\left(-4.6046\text{nJ}\right)\\ =12.8534\text{nJ}\\ \cong 12.8\text{nJ}\end{array}$

Conclusion:

Thus, the total energy stored in the system at $t=0\text{ms}$ and $t=5\text{ms}$ when a-b terminals are short-circuited in the circuit are $\underset{\_}{44.1\text{nJ}}$ and $\underset{\_}{12.8\text{nJ}}$, respectively.