Chapter 16, Problem 1P

Find the total impedance of the parallel networks of Fig. 1663 in rectangular and polar form.

Fig. 1663

To determine

(a)

Total impedance of the given network.

Answer to Problem 1P

Rectangular form: $Z_T=\left(941.17+j235.29\right)\text{Ω}$

Polar form: $Z_T=970.14\text{Ω}\angle \text{14}\text{.03}°$

Explanation of Solution

Given:

The given network is:

  

Calculation:

We can see from the given circuit that the resistor $R$ and the inductive reactance $X_L$ are in parallel. Therefore, the total resistance of the network will be:

  $\begin{array}{l}{Z}_T=R\parallel \left(j{X}_L\right)\\ Z_T=\frac{\left(1000\right)\left(j4000\right)}{1000+j4000}\\ Z_T=\frac{j4000000}{1000+j4000}\\ Z_T=\left(941.17+j235.29\right)\text{Ω}\\ Z_T=970.14\text{Ω}\angle \text{14}\text{.03}°\end{array}$

To determine

(b)

Total impedance of the given network.

Answer to Problem 1P

Rectangular form: $Z_T=\left(5679.23-j1153.85\right)\text{Ω}$

Polar form: $Z_T=5883.48\text{Ω}\angle -\text{11}\text{.31}°$

Explanation of Solution

Given:

The given network is:

  

Calculation:

We can see from the given circuit that the resistor $R$, capacitive reactance $X_C$ and the inductive reactance $X_L$ are in parallel. Therefore, the total resistance of the network will be:

  $\begin{array}{l}{Z}_T=R\parallel \left(j{X}_L\right)\parallel \left(-j{X}_C\right)\\ Z_T=\frac{1}{\frac{1}{j{X}_L}+\frac{1}{-j{X}_C}+\frac{1}{R}}\\ Z_T=\frac{1}{\frac{1}{20000\angle 90°}+\frac{1}{12000\angle -90°}+\frac{1}{6000}}\\ Z_T=\frac{1}{50\times {10}^{-6}\angle -90°+83.33\times {10}^{-6}\angle 90°+166.67\times {10}^{-6}}\\ Z_T=\left(5679.23-j1153.85\right)\text{Ω}\\ Z_T=5883.48\text{Ω}\angle -\text{11}\text{.31}°\end{array}$

To determine

(c)

Total impedance of the given network.

Answer to Problem 1P

Rectangular form: $Z_T=j1224.32\text{Ω}$

Polar form: $Z_T=1224.32\text{Ω}\angle \text{90}°$

Explanation of Solution

Given:

The given network is:

  

Calculation:

We can see from the given circuit that two inductors and one capacitor are connected in parallel.

Therefore, we need to calculate their equivalent reactance first:

Inductive reactance of $\text{40mH}$ inductor:

  $\begin{array}{l}{\text{X}}_{L_1}=2\pi f{L}_1\\ {\text{X}}_{L_1}=\left(2\pi \right)\left(10\times {10}^3\right)\left(40\times {10}^{-3}\right)\\ {\text{X}}_{L_1}=2513.27\text{Ω}\end{array}$

Inductive reactance of $\text{20mH}$ inductor:

  $\begin{array}{l}{\text{X}}_{L_1}=2\pi f{L}_2\\ {\text{X}}_{L_1}=\left(2\pi \right)\left(10\times {10}^3\right)\left(20\times {10}^{-3}\right)\\ {\text{X}}_{L_1}=1256.64\text{Ω}\end{array}$

Capacitive reactance of $6\text{nF}$ inductor:

  $\begin{array}{l}{\text{X}}_C=\frac{1}{2\pi fC}\\ {\text{X}}_C=\frac{1}{\left(2\pi \right)\left(10\times {10}^3\right)\left(6\times {10}^{-9}\right)}\\ {\text{X}}_C=2652.58\text{Ω}\end{array}$

Therefore, the total resistance of the network will be:

  $\begin{array}{l}{Z}_T=\left(j{X}_{L_1}\right)\parallel \left(j{X}_{L_2}\right)\parallel \left(-j{X}_C\right)\\ Z_T=\frac{1}{\frac{1}{j{X}_{L_1}}+\frac{1}{j{X}_{L_2}}+\frac{1}{-j{X}_C}}\\ Z_T=\frac{1}{\frac{1}{2513.27\angle 90°}+\frac{1}{1256.63\angle 90°}+\frac{1}{2652.58\angle -90°}}\\ Z_T=\frac{1}{398\times {10}^{-6}\angle -90°+377\times {10}^{-6}\angle 90°+795.78\times {10}^{-6}\angle -90°}\\ Z_T=\frac{1}{816.78\times {10}^{-6}\angle -90°}\\ Z_T=j1224.32\text{Ω}\\ Z_T=1224.32\text{Ω}\angle \text{90}°\end{array}$