Chapter 10, Problem 1P

Given the network in Fig. P10.1,

(a) write the equations for ${\upsilon }_a\left(t\right)\text{and}{\upsilon }_b\left(t\right)$.

(b) write the equations for ${\upsilon }_c\left(t\right)\text{and}{\upsilon }_d\left(t\right)$.

To determine

The equations for $v_a\left(t\right)\text{ and }{v}_b\left(t\right)$ for the given circuit.

Answer to Problem 1P

The equations are:

  $\begin{array}{l}{v}_a\left(t\right)=L_1\frac{d{i}_1\left(t\right)}{dt}-M\frac{d{i}_2}{dt}\\ v_b\left(t\right)=-L_2\frac{d{i}_2}{dt}-M\frac{d{i}_1}{dt}\end{array}$

Explanation of Solution

Given:

The given circuit is shown below.

Calculation:

Referring to the given circuit, applying the Kirchhoff’s voltage law to the left side of the loop:

  $\begin{array}{l}-v_a\left(t\right)+L_1\frac{d{i}_1\left(t\right)}{dt}-M\frac{d{i}_2\left(t\right)}{dt}=0\\ v_a\left(t\right)=L_1\frac{d{i}_1\left(t\right)}{dt}-M\frac{d{i}_2}{dt}\mathrm{..........}\left(1\right)\end{array}$

From the circuit:

  $v_a\left(t\right)=-v_c\left(t\right)\mathrm{..........}\left(2\right)$

Solving equation 1 and 2:

  $v_a\left(t\right)=-v_c\left(t\right)=L_1\frac{d{i}_1\left(t\right)}{dt}-M\frac{d{i}_2}{dt}$

Now applying the KVL to the right side of the loop:

  $\begin{array}{l}-v_b\left(t\right)-L_2\frac{d{i}_2}{dt}-M\frac{d{i}_1}{dt}=0\\ v_b\left(t\right)=-L_2\frac{d{i}_2}{dt}-M\frac{d{i}_1}{dt}\mathrm{............}\left(3\right)\end{array}$

Therefore, the equation will be:

  $\begin{array}{l}{v}_a\left(t\right)=L_1\frac{d{i}_1\left(t\right)}{dt}-M\frac{d{i}_2}{dt}\\ v_b\left(t\right)=-L_2\frac{d{i}_2}{dt}-M\frac{d{i}_1}{dt}\end{array}$

To determine

The equations for $v_c\left(t\right)\text{ and }{v}_d\left(t\right)$ for the given circuit.

Answer to Problem 1P

The equations are:

  $\begin{array}{l}{v}_c\left(t\right)=-L_1\frac{d{i}_1\left(t\right)}{dt}+M\frac{d{i}_2}{dt}\\ v_d\left(t\right)=L_2\frac{d{i}_2}{dt}+M\frac{d{i}_1}{dt}\end{array}$

Explanation of Solution

Given:

The given circuit is shown below.

Calculations:

From part (a), the equations of the $v_a\left(t\right)\text{ and }{v}_b\left(t\right)$ are:

  $\left\{\begin{array}{l}{v}_a\left(t\right)=L_1\frac{d{i}_1\left(t\right)}{dt}-M\frac{d{i}_2}{dt}\\ v_b\left(t\right)=-L_2\frac{d{i}_2}{dt}-M\frac{d{i}_1}{dt}\end{array}\right\}\mathrm{..........}\left(1\right)$

From the circuit:

  $\left\{\begin{array}{l}{v}_a\left(t\right)=-v_c\left(t\right)\\ v_b\left(t\right)=-v_d\left(t\right)\end{array}\right\}\mathrm{...............}\left(2\right)$

Hence, solving the equation:

  $\begin{array}{l}{v}_a\left(t\right)=-v_c\left(t\right)=L_1\frac{d{i}_1\left(t\right)}{dt}-M\frac{d{i}_2}{dt}\\ v_b\left(t\right)=-v_d\left(t\right)=-L_2\frac{d{i}_2}{dt}-M\frac{d{i}_1}{dt}\end{array}$

Therefore, the equations are:

  $\begin{array}{l}{v}_c\left(t\right)=-L_1\frac{d{i}_1\left(t\right)}{dt}+M\frac{d{i}_2}{dt}\\ v_d\left(t\right)=L_2\frac{d{i}_2}{dt}+M\frac{d{i}_1}{dt}\end{array}$