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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 ismeasured in volts, resistance in ohms, and current in amperes.) Find the resistance to electrical flow in a circuitthat has a voltage V = (40+30i) volts and current I = (-5+ 3i) amps._+_i/_Note: Answer in the forma + bi/c. If b is negative make sure to put a negative sign in the answer box.

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 class=
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 class=
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 class=

Respuesta :

we have the formula

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

substitute given values

[tex]R=\frac{40+30i}{-5+3i}[/tex]

Remember that

To divide complex numbers, multiply both the numerator and denominator by the conjugate of the denominator

the conjugate of the denominator is (-5-3i)

so

[tex]\begin{gathered} R=\frac{40+30\imaginaryI}{-5+3\imaginaryI}*\frac{-5-3i}{-5-3i}=\frac{-40(5)-40(3i)-30i(5)-30i(3i)}{25-9i^2}=\frac{-200-120i-150i-90i^2}{25-9(-1)}=\frac{-110-270i}{34} \\ \\ R=\frac{-110-270\imaginaryI}{34} \\ simplify \\ R=\frac{-55-135\imaginaryI}{17} \end{gathered}[/tex]

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