Chapter 9, Problem 68SGA

To determine

Find the current in the given circuit.

Answer to Problem 68SGA

The current of the circuit is $26.8+11.469j$ .

Explanation of Solution

Given:

Thevoltage of the circuit is $50+180j$ and impedance is $4+5j$ ohms.

Calculation:

Convert complex number in polar form.

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ \theta =\arctan \frac{b}{a},\text{when}a>0\\ \theta =\arctan \frac{b}{a}+\pi ,\text{when}a<0\end{array}$

For voltage

  $\begin{array}{l}a=50,b=180\\ r=\sqrt{{50}^2+{180}^2}=186.815\\ \theta =\arctan \frac{180}{50}=\arctan \frac{18}{5}=1.2998\end{array}$

For impedance

For voltage

  $\begin{array}{l}a=4,b=5\\ r=\sqrt{4^2+5^2}=6.4\\ \theta =\arctan \frac{5}{4}=0.896\end{array}$

Use formula $E=I\cdot Z$ .

  $\Rightarrow I=\frac{E}{Z}$

  $\begin{array}{l}\Rightarrow I=\frac{186.815\left[\cos \left(1.2998\right)+j\sin \left(1.2998\right)\right]}{6.4\left[\cos \left(0.896\right)+j\sin \left(0.896\right)\right]}\\ =\frac{186.815}{6.4}\left[\cos \left(1.2998-0.896\right)+j\sin \left(1.2998-0.896\right)\right]\\ =29.1898\left[\cos \left(0.4038\right)+j\sin \left(0.4038\right)\right]\\ =26.8+11.469j\end{array}$

Hence the current of the circuitis $26.8+11.469j$ .