Voltage input in a circuit is V = 300sin(omegat)V=300sin(ωt) with current I = 100cos(omegat)I=100cos(ωt). Average power loss in the circuit is??

1 Answer
Aug 9, 2017

There is no real power dissipated by the impedance.

Explanation:

Please observe that

100cos(omegat) = 100sin(omegat-pi/2)100cos(ωt)=100sin(ωtπ2)

this means the current is phase shifted +pi/2+π2 radians from the voltage.

We can write the voltage and current as magnitude and phase:

V = 300angle0V=3000
I = 100anglepi/2I=100π2

Solving the impedance equation:

V = IZV=IZ

for Z:

Z = V/IZ=VI

Z = (300angle0)/(100anglepi/2)Z=3000100π2

Z = 3angle-pi/2Z=3π2

This means that the impedance is an ideal 3 Farad capacitor.

A purely reactive impedance consumes no power, because it returns all of the energy on negative part of the cycle, that was introduced on the positive part of the cycle.