Chapter A.7, Problem 86AYU

To determine

To Find:The impedance of a circuit.

Answer to Problem 86AYU

  $Z=\frac{7}{4}+\frac{1}{4}i\Omega$

Explanation of Solution

Given: In n ac circuit with two parallel pathways, the total impedance $Z$ , in ohms , satisfies the formula $\frac{1}{Z}=\frac{1}{Z_1}+\frac{1}{Z_2}$ , where $Z_1$ is the impedance of the first pathway and $Z_2$ is the impedance of the second pathway. Determine the total impedance if the impedance of two pathways are $Z_1=2+i\Omega$ and $Z_2=4-3i\Omega$

Given:

  $\frac{1}{Z}=\frac{1}{Z_1}+\frac{1}{Z_2}\mathrm{...}(i)$

  $Z_1=2+i\Omega$

  $Z_2=4-3i\Omega$

First we calculate $\frac{1}{Z_1}$ and $\frac{1}{Z_2}$ by multiplying their numerator and denominator by conjugates of $2+i(i.e2-i)$ and $4-3i(i.e.4+3i)$ .

Then

  $\begin{array}{l}\frac{1}{Z_1}=\frac{1}{2+i}=\frac{1}{2+i}\cdot \frac{2-i}{2-i}=\frac{2-i}{4-i}=\frac{2-i}{4-(-1)}=\frac{2-i}{4+1}\\ =\frac{2-i}{5}\\ \frac{1}{Z_2}=\frac{1}{4-3i}=\frac{1}{4-3i}\cdot \frac{4+3i}{4+3i}=\frac{4+3i}{16-(-9)}=\frac{4+3i}{25}\end{array}$

Now

In equation (i)

  $\begin{array}{l}\frac{1}{Z}=\frac{1}{Z_1}+\frac{1}{Z_2}\\ \frac{1}{Z}=\frac{2-i}{5}+\frac{4+3i}{25}\\ \frac{1}{Z}=\frac{5(2-i)+(4+3i)}{25}\\ \frac{1}{Z}=\frac{10-5i+4+3i}{25}\\ \frac{1}{Z}=\frac{14-2i}{25}\\ Z=\frac{25}{14-2i}\end{array}$

Multiply numerator and denominator by conjugate value of the denominator

We get

  $\begin{array}{l}Z=\frac{25}{14-2i}\cdot \frac{14+2i}{14+2i}\\ Z=\frac{25(14+2i)}{196-{(2i)}^2}\\ Z=\frac{25(14+2i)}{196-(-4)}{i}^2=-1\\ Z=\frac{25(14+2i)}{200}\\ Z=\frac{14+2i}{8}=\frac{7+i}{4}\\ Z=\frac{7}{4}+\frac{1}{4}i\Omega \end{array}$

Hence, the total impedance in the circuit is $Z=\frac{7}{4}+\frac{1}{4}i\Omega$