# Voltage source -> Ideal switch -> Resistance = Reactive power

1. Jul 31, 2009

### foges

This is a problem i just cant get my head around.

When you have a circuit consisting of a "Ideal Voltage source -> Ideal switch -> Ideal Resistance" all in series. The voltage source produces a sinusodal voltage and the switch is set to cut off a part of any rising curve:

http://img383.imageshack.us/img383/8819/picture2ixx.png [Broken]

Now if you do the calculations for the effective power that is dissapated at the resistance and the apparent power given off by the voltage source you will see that the apparent power is quite a bit higher than the effective power (P = 0.89*S for alpha = pi/3, (see picture)).

$$P_R = I_{eff}^2\cdot R, S = I_{eff}\cdot U_{eff}$$

How does this happen? where is the reactive power going?

PS: I have done all the calculations, im not asking for those, im looking for an explanation. Thanks :)

Last edited by a moderator: May 4, 2017
2. Jul 31, 2009

### vk6kro

The instantaneous power in a resistor is equal to the instantaneous voltage times the instantaneous current. These occur at the same time so there is no reactive component.

3. Jul 31, 2009

### foges

That would seem logical (what i thought), but check out the calculations (its in german, but shouldnt make much of a difference):

http://img509.imageshack.us/img509/4929/picture4w.png [Broken]

Last edited by a moderator: May 4, 2017
4. Jul 31, 2009

### vk6kro

Lots of luck with that......

The voltage across the resistor is the same as the voltage from the source when the switch is closed.
When the switch is open no power goes anywhere.

This is simple Ohms Law stuff. There is a voltage across a resistor and a current flowing in it. Why would there be any reactive component?