Chapter 19, Problem 1P

Using supeerposition, determine the current through the inductance X_L for the network of Fig. 19.105.

To determine

The current flowing through the inductance $X_L$ of the given circuit.

Answer to Problem 1P

The current through the inductor is ${\text{I}}_{\text{L}}=6.1\angle -32.11°\text{A}$.

Explanation of Solution

Given:

The given circuit is shown in Figure 1.

Calculation:

To apply superposition theorem, first consider the effect of voltage source ${\text{E}}_{\text{1}}$ and replace the source ${\text{E}}_{\text{2}}$ by a short circuit.

The required diagram is shown in Figure 2.

Let the voltage at node 1 is $V$. Apply KCL at node 1.

  $\begin{array}{l}\frac{V-30\angle 30°}{3}+\frac{V}{j8}+\frac{V}{-j6}=0\\ \left(\frac{1}{3}+\frac{1}{j8}+\frac{1}{-j6}\right)V=\frac{30\angle 30°}{3}\\ \left(0.333+j0.0417\right)=10\angle 30°\\ V=29.77\angle 22.87°\text{V}\end{array}$

The current through the inductor is given by

  ${\text{I}}_{\text{L1}}=\frac{V}{j8}$

Substitute $29.77\angle 22.87°$ for $V$ in the above equation.

  $\begin{array}{c}{\text{I}}_{\text{L1}}=\frac{29.77\angle 22.87°}{j8}\\ =3.72\angle -67.12°\text{A}\end{array}$

Now consider the effect of source voltage ${\text{E}}_{\text{2}}$ and replace the source voltage ${\text{E}}_{\text{1}}$ by short circuit.

The required diagram is shown in Figure 3.

Apply the KCL at node 2.

  $\begin{array}{c}\frac{V}{3}+\frac{V}{j8}+\frac{V-60\angle 10°}{-j6}=0\\ \left(\frac{1}{3}+\frac{1}{j8}+\frac{1}{-j6}\right)V=\frac{60\angle 10°}{-j6}\\ \left(0.333+j0.0417\right)=10\angle 100°\\ V=29.77\angle 92.87°\text{V}\end{array}$

The current through the inductor is given by,

  $\begin{array}{c}{\text{I}}_{\text{L2}}=\frac{29.77\angle 92.87°}{j8}\\ =3.72\angle 2.87°\text{A}\end{array}$

Therefore, according to super position theorem the current through the inductor is given by

  ${\text{I}}_{\text{L}}={\text{I}}_{\text{L1}}+{\text{I}}_{\text{L2}}$

Substitute $3.72\angle -67.12°\text{A}$ for ${\text{I}}_{\text{L1}}$ and $3.72\angle 2.87°\text{A}$ for ${\text{I}}_{\text{L2}}$ in the above equation.

  $\begin{array}{c}{\text{I}}_{\text{L}}=3.87\angle -67.12°+3.87\angle 92.87°\\ {\text{I}}_{\text{L}}=6.1\angle -32.11°\text{A}\end{array}$

Conclusion:

Therefore, the current through the inductor is $\text{I}=6.1\angle -32.11°\text{A}$.