Chapter 17, Problem 1P

For theseries-parallel network in Fig.17.38.

a. Calculate Z_T.

b. Determine I_s.

c. Determine I₁.

d. Find I₂.

e. Find V_L._.

Fig.17.38

To determine

(a)

Total impedance $Z_T$.

Answer to Problem 1P

  $Z_T=4\Omega \angle -22.65°$

Explanation of Solution

Given:

The given circuit is:

  

Calculation:

Let us modify the circuit as below to calculate the total impedance.

  

In the modified diagram,

Impedance $Z_1$ is nothing but the inductive reactance $X_L$ which is equal to $4\Omega$.

Impedance $Z_2$ is nothing but the parallel combination of $R$ and $X_C$.

Therefore, total impedance will be:

  $\begin{array}{l}{Z}_T=Z_1+Z_2\\ \text{where,}\\ Z_1=X_L\angle 90°\\ Z_1=4\Omega \angle 90°\\ Z_1=j4\Omega \\ \text{and}\\ {\text{Z}}_2=\frac{\left(X_C\angle -90°\right)\left(R\angle 0°\right)}{-j{X}_C+R}\\ {\text{Z}}_2=\frac{\left(8\Omega \angle -90°\right)\left(12\Omega \angle 0°\right)}{-j8\Omega +12\Omega }\\ {\text{Z}}_2=\frac{96\Omega \angle -90°}{12\Omega -j8\Omega }\\ {\text{Z}}_2=\frac{96\Omega \angle -90°}{14.42\Omega \angle -33.69°}\\ {\text{Z}}_2=6.657\Omega \angle -56.31°\\ {\text{Z}}_2=3.69\Omega -j5.54\Omega \end{array}$

Putting the values for the total impedance:

  $\begin{array}{l}{Z}_T=Z_1+Z_2\\ Z_T=\left(j4\Omega \right)+\left(3.69\Omega -j5.54\Omega \right)\\ Z_T=3.69\Omega -j1.54\Omega \\ Z_T=4\Omega \angle -22.65°\end{array}$

To determine

(b)

The current $I_S$.

Answer to Problem 1P

  $I_S=3.5\text{A}\angle 22.65°$

Explanation of Solution

Given:

The given circuit is:

  

Calculation:

Let us modify the circuit as below to calculate the total impedance.

  

In the modified diagram,

Impedance $Z_1$ is nothing but the inductive reactance $X_L$ which is equal to $4\Omega$.

Impedance $Z_2$ is nothing but the parallel combination of, $R$ and $X_C$.

Current $I_S$ :

  $\begin{array}{l}{I}_S=\frac{E}{Z_T}\\ I_S=\frac{\text{14V}\angle 0°}{4\Omega \angle -22.65°}\\ I_S=3.5\text{A}\angle 22.65°\end{array}$

To determine

(c)

The current $I_1$.

Answer to Problem 1P

  $I_1=3.5\text{A}\angle 22.65°$

Explanation of Solution

Given:

The given circuit is:

  

Calculation:

Let us modify the circuit as below to calculate the total impedance.

  

In the modified diagram,

Impedance $Z_1$ is nothing but the inductive reactance $X_L$ which is equal to $4\Omega$.

Impedance $Z_2$ is nothing but the parallel combination of $R$ and $X_C$.

Current $I_1$ :

We can see from the circuit that,

  $\begin{array}{l}{I}_1=I_S\\ I_1=3.5\text{A}\angle 22.65°\end{array}$

To determine

(d)

The current $I_2$.

Answer to Problem 1P

  $I_2=1.94\text{A}\angle -33.66°$

Explanation of Solution

Given:

The given circuit is:

  

Calculation:

Let us modify the circuit as below to calculate the total impedance.

  

In the modified diagram,

Impedance $Z_1$ is nothing but the inductive reactance $X_L$ which is equal to $4\Omega$.

Impedance $Z_2$ is nothing but the parallel combination of $R$ and $X_C$.

Current $I_2$ :

  $\begin{array}{l}{I}_2=\left(\frac{X_C\angle -90°}{X_C\angle -90°+R\angle 0°}\right)I_S\\ I_2=\left(\frac{8\Omega \angle -90°}{8\Omega \angle -90°+12\Omega \angle 0°}\right)\left(3.5\text{A}\angle 22.65°\right)\\ I_2=\left(\frac{8\Omega \angle -90°}{14.42\Omega \angle -33.69°}\right)\left(3.5\text{A}\angle 22.65°\right)\\ I_2=\left(0.554\angle -56.31°\right)\left(3.02\text{A}\angle -37.17°\right)\\ I_2=1.94\text{A}\angle -33.66°\end{array}$

To determine

(e)

Voltage $V_L$.

Answer to Problem 1P

  $V_L=14\text{V}\angle 112.65°$

Explanation of Solution

Given:

The given circuit is:

  

Calculation:

Let us modify the circuit as below to calculate the total impedance.

  

In the modified diagram,

Impedance $Z_1$ is nothing but the inductive reactance $X_L$ which is equal to $4\Omega$.

Impedance $Z_2$ is nothing but the parallel combination of $R$ and $X_C$.

Voltage $V_L$ :

  $\begin{array}{l}{V}_L=I_S{Z}_1\\ V_L=\left(3.5\text{A}\angle 22.65°\right)\left(4\Omega \angle 90°\right)\\ V_L=14\text{V}\angle 112.65°\end{array}$