=Complex numbers may be applied to electrical circuits. Electrical engineers use the fact that resistance R toelectrical flow of the electrical current I and the voltage V are related by the formula V RI. (Voltage is measured in volts, resistance in ohms, and current in amperes.) Find the resistance to electrical flow in a circuit(90+20i) volts and current I = (- 5+3i) amps._+_/_Note: Answer in the forma + bi/c. If b is negative make sure to put a negative sign in the answer box.

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PhineusA33421 PhineusA33421 · Answer 1 · 2022-10-30T22:01:00+01:00

Since the given formula is

[tex]V=RI[/tex]

Since we need to find R, then divide both sides by I

[tex]\begin{gathered} \frac{V}{I}=\frac{RI}{I} \\ \\ \frac{V}{I}=R \\ \\ R=\frac{V}{I} \end{gathered}[/tex]

Since V = (90 + 20i)

Since I = (-5 + 3i), then

[tex]R=\frac{(90+20i)}{(-5+3i)}[/tex]

Multiply up and down by the conjugate of (-5 + 3i) which is (-5 - 3i)

[tex]R=\frac{(90+20i)}{(-5+3i)}\times\frac{(-5-3i)}{(-5-3i)}[/tex]

Simplify up and down

[tex]R=\frac{(90)(-5)+(90)(-3i)+(20i)(-5)+(20i)(-3i)}{(-5)(-5)-(3i)(3i)}[/tex][tex]R=\frac{-450-270i-100i-60i^2}{25-9i^2}[/tex]

Replace i^2 by -1 and add the like terms

[tex]\begin{gathered} R=\frac{-450-370i-60(-1)}{25-9(-1)} \\ \\ R=\frac{-450-370i+60}{25+9} \\ \\ R=\frac{-390-370i}{34} \end{gathered}[/tex]

The answer is

[tex]R=\frac{[-390]+[-370]i}{34}[/tex][tex]R=\frac{[-195]+[-185]i}{17}[/tex]