Question

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

Answer 1

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.