# RC Circuit

Discussion in 'Homework Help' started by Hitman6267, May 4, 2010.

1. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0

I'm planning on using formula stated below to solve this.
Formula: $V(t) = V_f + (V_0 - V_f)e^{-\frac{t}{time constant}}$

which becomes i(time constant) = (i0 - if) e^(-t/timeconstant) + if

Ok so the first step is to calculate the time constant

time constant= L/Rth

To calculate Rth I'm going to short circuit the independent sources,
in another thread Ghar proved that R would be 2x (R1||R2). He proved it quite eloquently but if some one has a short way to prove it please post it.

From now on I'm going to use τ as time constant.
τ= 4.8x 10^-3 / 5.83 = 8.23x 10^-4

Next I need to calculate I0
So I calculated Req for the whole circuit and Got Isrc = 3.39
I used current divider to get the current that goes in the H like circuit on the right. I1=2.06.

I0 is going to be the difference between the current that goes through R1 and the current that goes through R2.

I'm not sure how to use current divider here. Any one can help ?

Also can't see how calculating I final would be different. If i replace the inductor by a short circuit that doesn't effect the Req so I'll get the same answer.

Last edited: May 4, 2010
2. ### Ghar Active Member

Mar 8, 2010
655
73
Initial inductor current is zero because there was no source connected for a long time.
Final inductor current will be the short circuit current because after a long time an inductor is a short.

Current division here is annoying because of all the parallel combinations. It's easier to find the DC voltage at A (equal to B) which is 6V by symmetry. With that voltage you can easily get the current.

I'm not sure where you got a current of 2.06, I don't see that number anywhere.
3.39 mA is correct however. Make sure you write down mA though... don't want to keep losing your orders of magnitude.

Edit:
Nevermind, I see the 2.06 now.

Last edited: May 5, 2010
3. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,397
355
This circuit lends itself to solution by the loop method.

First of all, get rid of the 9kΩ resistor, R. It's connected across a voltage source and though it conducts a current, that current is confined to the small loop between the source and R; it has no effect on the problem.

Choose 3 clockwise loop currents like this:

I1 is the loop consisting of the 12 volt source and the leftmost two resistors of the "H" bridge; it follows the path labeled i1 passing through the top left R1.

I2 is the loop which goes clockwise all the way around the outside of the "H" bridge.

I3 is the loop which goes clockwise around the bottom mesh of the "H" bridge. I3 will be the final current in the inductor after many time constants have passed.

With this choice of loop currents, we have I1 as the total current into the bridge (and out of the source); I3 is the current through the inductor (which is a short between terminals a and b).

The current already labeled i1 on the schematic will be given by (I1-I2). The current downward in the top right resistor will be just I2.

$\left[ \begin{array}{3}12000 & -12000 & -7000\\-12000 & 24000 & 12000\\-7000 & 12000 & 12000\end{array}\right]\left[ \begin{array}{1}I_1\\I_2\\I_3\\\end{array}\right]=\left[ \begin{array}{1}12\\0\\0\\\end{array}\right]$

Do you know how to solve a system like this? You will get I3 directly and with no need to find Vth or Rth, although you can find them from this system.

To help you know if you've got it right, I will tell you that I1 is 2.057 mA; that's the total current from the 12 volt source.

Since the problem (apparently) asks for the current in the inductor after 1 time constant, you don't even need to calculate the time constant. In a single energy circuit like this where the current follows a single exponential law, and where the initial value is zero, the value after 1 time constant is:

$I(\tau) = I_f (1-e^{-1})= .632*I_f=.632*I3$

Last edited: May 5, 2010