Chapter 9.5, Problem 36E

(a)

To determine

To find: the complex numbers that represent the impedances in the two parts of the circuit.

Answer to Problem 36E

  $Z_1=3$ .

Explanation of Solution

Given:

  $z=R+\left(X_L-X_C\right)j$ .

Concept used:

As opposed to constant voltage circuits, in AC circuits the impedance of an element is a measure of how much the element opposes current flow when an AC voltage is applied across it. It is basically a voltage to current ratio, expressed in the frequency domain.

Impedance is a complex number which consist of a real and an imaginary part.

  $Z=R+jX$ .

Where Z is the complex impedance. The real part R represents resistance, while the imaginary X represents reactance.

Calculation:

Consider the impedance as:

  $z=R+\left(X_L-X_C\right)j$ .

In the first part of the circuit.

Reactance due to capacitance $X_c=2\text{ ohms}$ .

Reactance due to inductance $X_L=1\text{ ohms}$ .

In the second part of the circuit.

Resistance $R=3\text{ ohms}$ .

Reactance due to capacitance $X_c=1\text{ ohm}$ .

Reactance due to inductance $X_L=1\text{ ohm}$ .

Complex number that represent impedance $\left(Z\right)$ in the first part of the circuit is:

  $\begin{array}{l}{Z}_1=R+\left(X_L-X_C\right)j.\\ \Rightarrow 10+\left(1-2\right)j.\\ \Rightarrow 10-j.\end{array}$

Complex number that represent impedance $\left(Z\right)$ in the second part of the circuit is:

  $\begin{array}{l}{Z}_1=R+\left(X_L-X_C\right)j.\\ \Rightarrow 3+\left(1-1\right)j.\\ \Rightarrow 3.\end{array}$

Hence, $Z_1=3$ .

(b)

To determine

To find: the total impedance in the circuit.

Answer to Problem 36E

Total impedance is $13-j$ .

Explanation of Solution

Given:

  $z=R+\left(X_L-X_C\right)j$ .

Concept used:

As opposed to constant voltage circuits, in AC circuits the impedance of an element is a measure of how much the element opposes current flow when an AC voltage is applied across it. It is basically a voltage to current ratio, expressed in the frequency domain.

Impedance is a complex number which consist of a real and an imaginary part.

  $Z=R+jX$ .

Where Z is the complex impedance. The real part R represents resistance, while the imaginary X represents reactance.

Calculation:

Total impedance $\left(Z\right)$ :

  $\begin{array}{l}Z=Z_1+Z_2.\\ \Rightarrow 10-j+3.\\ \Rightarrow 13-j.\end{array}$

Hence, total impedance is $13-j$ .

(c)

To determine

To find: the admittance in a circuit with an impedance $6+3j\text{ ohms}$ .

Answer to Problem 36E

Admittance is $\frac{2}{15}-\frac{j}{15}$ .

Explanation of Solution

Given:

  $z=R+\left(X_L-X_C\right)j$ .

Concept used:

As opposed to constant voltage circuits, in AC circuits the impedance of an element is a measure of how much the element opposes current flow when an AC voltage is applied across it. It is basically a voltage to current ratio, expressed in the frequency domain.

Impedance is a complex number which consist of a real and an imaginary part.

  $Z=R+jX$ .

Where Z is the complex impedance. The real part R represents resistance, while the imaginary X represents reactance.

Calculation:

Now the admittance $\left(S\right)$ :

  $\begin{array}{l}S=\frac{1}{Z}.\\ \Rightarrow \frac{1}{6+3j}.\end{array}$

Rationalizing the equation:

  $\begin{array}{l}\frac{1}{6+3j}.\frac{6-3j}{6-3j}.\\ \Rightarrow \frac{6-3j}{36+9}.\\ \Rightarrow \frac{2}{15}-\frac{j}{15}.\end{array}$

Hence, admittance is $\frac{2}{15}-\frac{j}{15}$ .