Engineering Mathematics?

Question

If the current is given by the voltage 4.50 - j7.50mA and the impedance is 7.10 - j2.15kΩ, find the magnitude of the voltage.

( j = √-1)

Answer 1

Use Ohm's Law(V=IZ) where Z is the impedance, multiply them out. You'll get another A+jB equation. To find the Magnitude you take the squareroot(A^2+B^2) and to get the phase angle you'll take arctan(B/A).

If I did the calculations correctly then

(4.5-j7.5mA)*(7.5-j2.15)kohms

=15.825-j62.925

sqrt(15.825^2+(-62.925)62) = 64.88

arctan(-62.925/15.825) = -14.387

therefore your magnitude is 64.88V and your phase angle is -14.387

I hope this helps