Question

electricity what is the voltage v of an electronic circuit with a current c of 5 - 2j and an impedance i of 4 + j? use the formula v = ci.
a) 19 - j
b) 18 - 3j
c) 20 - 2j
d) 22 - 3j

Explanation:

Step1: Multiply complex - numbers

Given $C = 5-2j$ and $I = 4 + j$, use the formula $V=CI=(5 - 2j)(4 + j)$.
Apply the FOIL method: $(a + b)(c + d)=ac+ad+bc+bd$. Here, $a = 5$, $b=-2j$, $c = 4$, $d = j$.
$V=(5\times4)+(5\times j)+(-2j\times4)+(-2j\times j)$.

Step2: Simplify each term

$5\times4 = 20$, $5\times j=5j$, $-2j\times4=-8j$, and $-2j\times j=-2j^{2}$.
Since $j^{2}=-1$, then $-2j^{2}=-2\times(-1) = 2$.
So $V=20 + 5j-8j+2$.

Step3: Combine like - terms

Combine the real parts ($20$ and $2$) and the imaginary parts ($5j$ and $-8j$).
The real - part sum is $20 + 2=22$, and the imaginary - part sum is $5j-8j=-3j$.
So $V = 22-3j$.

Answer:

D. $22 - 3j$