Question

3 - 3 application of complex numbers tai alison. complex numbers are used with electricity. in these problems, j usually represents the imaginary unit. in a circuit, alternating current, the voltage, and impedance (or hindrance) to current can be represented by complex numbers. ex: in an ac circuit, the voltage v, current c, and impedance i are related by the formula v = c · i. find the voltage in a circuit with current 2 + 4j amps and impedance 9 - 3j ohms.

Explanation:

Step1: Recall the formula

We are given the formula $V = C\cdot I$. We know that $I = 2 + 4j$ (current) and $C=9 - 3j$ (impedance).

Step2: Substitute values

Substitute the values of $I$ and $C$ into the formula: $V=(9 - 3j)(2 + 4j)$.

Step3: Expand using FOIL

$(9 - 3j)(2 + 4j)=9\times2+9\times4j-3j\times2-3j\times4j$.
$=18 + 36j-6j - 12j^{2}$.

Step4: Simplify using $j^{2}=- 1$

Since $j^{2}=-1$, we have $18 + 36j-6j-12\times(-1)$.
$=18 + 36j-6j + 12$.

Step5: Combine like - terms

$(18 + 12)+(36j-6j)=30 + 30j$.

Answer:

$30 + 30j$