# RLC circuit

Discussion in 'Homework Help' started by Eng_Student, Sep 5, 2009.

1. ### Eng_Student Thread Starter New Member

Sep 5, 2009
2
0
Hi, I have a homework problem for a Power Electronics course that I cannot figure out how to do, except perhaps by differential equations and I don't think that's what the instructor intends...

Here's a really poor illustration:

+ VL -
+ _____000__________
L | | +
| <
Vi C = > R v(t)
| < °
- ____________|_____| -

I am given that an AC input voltage with T=6μs has a value of 15V for 4μs and 0V for 2μs. I am also told that L=5μH and that P, the power dissipation across the R load is 250W. I am asked to calculate the average output voltage. That is all the information I am given.

From the period T I was able to calculate a frequency of 166.6kHz and then ω=1.046Mrad/s.

After that, my approach was to find the total impedance Z, which is jωL + jωRC/(jωC + R). Then the total current is the input voltage divided by the total impedance. And the total current is equal to the sum of the capacitor current and the resistor current, which is C(dv/dt) + v/R, where v is the output voltage.

So now I have an equation that relates the input and output voltages. Once I calculate the ouput voltage, I can find its average and that is my answer. However I have the following problems:

a) This is a differential equation and I don't think our instructor intends for us to solve a differential equation. If he does, then I am not sure how to do it. Actually, I have a feeling there is a much simpler method to solve this problem which I cannot see.

b) I don't have values for R or C, and so I can't calculate any impedances, etc. Basically, it doesn't look like I have enough information to solve the problem.

If anyone can offer me some advice I would really appreciate it. Thanks.

2. ### Eng_Student Thread Starter New Member

Sep 5, 2009
2
0
All of the spaces got erased, so the drawing makes no sense. But basically, it was an inductor and then a capacitor and resistor in parallel.
The output voltage was over the parallel capacitor and resistor.