LC Parallele Voltmeter problem

1. Jul 8, 2010

Gizm0

http://img295.imageshack.us/img295/6016/55877904.th.jpg [Broken]

I tried everything to solve this.
XL and XC are an inductor and a capacitor (respectively).

I start with the idea that there are 50 V on each of them,
and that a current of 5 A flows between,
and that the generator has no effect,
because the potential on each end is the same.
But I can't figure out what current the ampermeter shows.
but i can't get it... tryed to replace the LC tree with a current generator, (duno how stupid the idea is)
and everything else that came to my mind...but I just don't know the trick.

Last edited by a moderator: May 4, 2017
2. Jul 9, 2010

stevenb

Perhaps there is a trick, but if you don't see a trick, then just use normal phasor analysis using complex impedances.

The basic approach is to first calculate the generator voltage using the fact that the voltmeter reads 100 V with the switch turned toward the voltmeter. When you do this calculation keep in mind that the inductive reactance has an implied j (i.e. square root of -1) and the capacitive reactance has an implied -j. These imaginary values are indications of 90 degree phase lead and 90 degree phase lag compared to a pure resistance.

It is straightforward to calculate the voltage using a voltage divider formula on each load branch. The voltmeter then reads the difference in voltage across each voltage division.

Once you know the generator voltage, you can calculate the ampmeter current after the switch is thrown to close that path. This requires a careful calculation, but is straightforward. Again remember that the capacitive reactance has in implied -j with it.

I tried to do the full problem out, but went rather quickly and didn't get the right answer. I did calculate the generator voltage as having a magnitude of 100 V, which I think is correct. There is a phase shift of 90 degrees as indicated by the result that V=j*U. However, the phase shift shouldn't be relevant here because the voltmeter and ampmeter will presumably only measure magnitude. When I tried to calculate the current I got about 3.33*sqrt(2), but I probably made a mistake and ended up with a 10/3 factor at one point, rather than a 10/4=2.5. Again, doing this final calculation requires taking the magnitude of a complex number (such as 1+j or 1-j, for example) which is where the sqrt(2) enters in. If it weren't 2:30 AM, I would find my mistake, but I'm too tired right now. I'm sure you can do it though.

3. Jul 9, 2010

Gizm0

Solved it,
I know that the reactances are (j10) and (-j10),
but I ignored the fact that the generator is actually a short circuit.

Because of that the inductor is parallel to R,
and the capacitor is also parallel with R.
It's just straight forward calculation and easy to solve.
The voltmeter displays the voltage of a Thevenin generator,
and the resulting impedance (parallele R and XL + parallele R and Xc) goes in series with the generator.