Chapter 9.6, Problem 48E

a.

To determine

To convert:

The two sources of impedance to complex numbers in rectangular form.

Answer to Problem 48E

The complex numbers are given by $7.65+6.44j$ and $14.04+7.67j$ .

Explanation of Solution

Given information:

The two sources are given by $10\left(\cos 0.7+j\sin 0.7\right)ohms$ and $16\left(\cos 0.5+j\sin 0.5\right)ohms$ .

Consider the two sources $10\left(\cos 0.7+j\sin 0.7\right)ohms$ and $16\left(\cos 0.5+j\sin 0.5\right)ohms$ .

Converting these sources to complex numbers in rectangular form:

  $10\left(\cos 0.7+j\sin 0.7\right)=10\cos 0.7+10j\sin 0.7\approx 7.65+6.44j$

  $16\left(\cos 0.5+j\sin 0.5\right)=16\cos 0.5+16j\sin 0.5\approx 14.04+7.67j$

Therefore, the complex numbers are given by $7.65+6.44j$ and $14.04+7.67j$ .

b.

To determine

To find:

The total impedance in the circuit.

Answer to Problem 48E

The total impedance is given by $21.69+14.11j$ .

Explanation of Solution

Given information:

The two sources are given by $10\left(\cos 0.7+j\sin 0.7\right)ohms$ and $16\left(\cos 0.5+j\sin 0.5\right)ohms$ .

From part (a) the complex numbers are given by $7.65+6.44j$ and $14.04+7.67j$

Now the total impedance is given by

  $7.65+6.44j+14.04+7.67j=21.69+14.11j$

Therefore, the total impedance is given by $21.69+14.11j$ .

c.

To determine

To convert:

The total impedance into polar form.

Answer to Problem 48E

The polar form is given by $25.88\left(\cos \left(0.58\right)+i\sin \left(0.58\right)\right)ohms$ .

Explanation of Solution

Given information:

The two sources are given by $10\left(\cos 0.7+j\sin 0.7\right)ohms$ and $16\left(\cos 0.5+j\sin 0.5\right)ohms$ .

Converting these complex numbers back to polar form:

  $r=\sqrt{{\left(21.69\right)}^2+{\left(14.11\right)}^2}=25.28$

  $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)={\tan }^{-1}\left(\frac{14.11}{21.69}\right)=0.58$

  $\begin{array}{l}21.69+14.11j=r\left(\cos \theta +i\sin \theta \right)\\ \Rightarrow 25.88\left(\cos \left(0.58\right)+i\sin \left(0.58\right)\right)\end{array}$

Therefore, the polar form is given by $25.88\left(\cos \left(0.58\right)+i\sin \left(0.58\right)\right)ohms$ .