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

1 Answer
Douglas K.
Aug 9, 2017

There is no real power dissipated by the impedance.

Explanation:

Please observe that

#100cos(omegat) = 100sin(omegat-pi/2)#

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

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

#V = 300angle0#
#I = 100anglepi/2#

Solving the impedance equation:

#V = IZ#

for Z:

#Z = V/I#

#Z = (300angle0)/(100anglepi/2)#

#Z = 3angle-pi/2#

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.

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