thevenins equivalent circuit

Thread Starter

Richie K

Joined Feb 7, 2004
Hi all, hope somebody can help me out here. I have a 1kohm resistor, 0.1uF capacitor and an AC supply of 56.57sin(10000t+20degrees)V all in one series loop. I need to find the thevenin's equivalent cct between terminals A and B, which is across the capacitor. I've worked out the thevenin's resistance, but am having trouble with the way the voltage is expressed. We have not been given a value for t, so I don't really see how I can work out V and therfore can't work out the thevenin's voltage. :wacko:


Joined Nov 14, 2003
The voltage source is, as you said, an ac supply. It's voltage depends on time, so you don't have to evaluate it at a specific time and set it's voltage to a single value.

Someone correct me if I'm wrong but isn't simplifying the circuit to it's Thevenin equivalent a procedure used to simplify a circuit comprised of resistors and voltage/current sources only. I don't remember doing this for any caps, or if I have then I ignored (opened) the caps.

It seems that the circuit is already in it's Thevenin equivalent form: Vth = 56.57sin(10000t+20degrees)V and Rth = 1k-ohm.

You aren't using Z=1/jwC to represent the impedance of the capacitor are you- this is not a bode plot problem, right?

Thread Starter

Richie K

Joined Feb 7, 2004
Thanks for the reply, I was thinking a little along your lines, I also only ever used voltage/current sources and resistors. This is why I'm having some difficulty, I wonder if it may be a mistake in the question. I did use Z=1/jwC for the cap, thats the only way I could see to make it work. I am not sure if maybe I should state a time at the beginning to work from and use it to give a voltage at that particular time.


Joined Nov 17, 2003
For a Thevenin Equivalent circuit, the series resitance can compose of resistive, capactitive and inductive elements. Therefore if you have a resistor and a capacitor, your Thevenin resistance will be a complex value. Work out R Thevenin asd you would any comlex impedance.

Thevenin voltage is merely 56.57sin(10000t+20degrees)V, the value of t only becomes important when you come to evaluate the circuit beyond the question you have asked.

From what I can gather, the information given from the volatge is to allow you to calculate the Capacitive reactance. Frequency = 10000/2*pi and the Capactive reactance Xc = 1/2*pi*f*C where C = 0.1uF. Xc = 1000 ohms (Check this though)

Post back if you are struggling :)


Joined Feb 17, 2004
The Thevenin impedance looking into the terminals 'a' and 'b' would be the the resistance R (1kohm) in parallel with the capacitance c (1uF). Since you know the operating frequency of the source (10000 radians/sec) you can calculate the complex impedance of the paralell network.

The Thevinin voltage looking in from terminals 'a' and 'b' would be the voltage drop across the capacitor. This can be computed by using a simple voltage divider formula. This would look something like this Vth=Vsource*(Zc/(R+Zc)). This calculation will involve complex numbers and is best worked out in polar coordinates.

Good Luck