Question

if the current in a circuit is 2 - j5 amps and the resistance is 1 + j3 ohms, what is the voltage?
a.) 17 + j volts
b.) 13 + j11 volts
c.) 17 + j11 volts
d.) 13 + j volts

Explanation:

Step1: Recall Ohm's Law

According to Ohm's Law $V = I\times R$, where $V$ is voltage, $I$ is current and $R$ is resistance. Here $I = 2 - j5$ and $R=1 + j3$.

Step2: Multiply complex - numbers

$(2 - j5)\times(1 + j3)=2\times(1 + j3)-j5\times(1 + j3)$.
Expand the right - hand side:
\[

$$\begin{align*} 2\times(1 + j3)-j5\times(1 + j3)&=2 + j6-(j5 + j^{2}15)\\ \end{align*}$$

\]
Since $j^{2}=-1$, we have:
\[

$$\begin{align*} 2 + j6-(j5-15)&=2 + j6 - j5 + 15\\ &=(2 + 15)+j(6 - 5)\\ &=17 + j \end{align*}$$

\]

Answer:

A. $17 + j$ volts