Finding Thevenin equivalent

I'm looking to find the Thevenin equivalent of two parallel voltage sources where each source has some internal resistance as well for simulation in Matlab. I don't know where and how exactly to put a voltage meter and current meter in place to measure the open source voltage and short circuit current. Can anyone guide me?

• We'd need a schematic, for a start – user17592 Apr 13 '13 at 8:28

OK, say we have an unknown circuit and we want to measure its Thevenin equivalent:

We have a resistive load $R_L$ connected to our circuit and we don't want to remove it while measuring the necessary quantities. First we measure the short-circuit current by connecting an ammeter between the measurement node and ground. The ammeter will short-circuit the load, which is exactly what we want in order to measure the short-circuit current. Let's call the result of our measurement $I_{Th}$. This current defines the ratio of $V_{Th}$ and $R_{Th}$: $$I_{Th}=\frac{V_{Th}}{R_{Th}} \tag{1}$$

Since we cannot remove the load, we cannot directly measure the open-circuit voltage $V_{Th}$. Instead, we measure the voltage $V_L$ across the load. The current through the load is then

$$I_L=\frac{V_L}{R_L}=\frac{V_{Th}}{R_{Th}+R_L} \tag{2}$$

Equations (1) and (2) allow us to solve for the unknown quantities $V_{Th}$ and $R_{Th}$:

$$V_{Th} = \frac{V_LR_LI_{Th}}{I_{Th}R_L-V_L}$$

$$R_{Th} = \frac{V_LR_L}{I_{Th}R_L-V_L}$$

• very well , I guess this is finally what i was looking for ! thank you very much for taking time to answer my questions . thanks a looooooooooooooooooooooooooot indeed – sahar Apr 14 '13 at 17:03
• Welcome, you could accept the answer to make it clear to everyone else that your question has been answered satisfactorily. – Matt L. Apr 14 '13 at 17:32
• oh how should i do that ? i'm quite new to this place . by sharing ? – sahar Apr 14 '13 at 17:34
• On the top left of the answer you can upvote (answer is useful), downvote (not useful) or accept (problem solved) an answer. – Matt L. Apr 14 '13 at 18:26
• oh that's nice too .i thought that 's a picture only , didn try clicking on that . thanks again – sahar Apr 14 '13 at 18:38

Let's assume there are two parallel voltage sources $V_1$ and $V_2$ with internal resistances $R_1$ and $R_2$:

To compute the Thevenin equivalent you need the open-circuit voltage $V_{TH}$ and the total internal resistance $R_{Th}$, or, equivalently, the short-circuit current $I_{Th}$. For $V_{Th}$ you obtain $$V_{Th} = \frac{V_1R_2+V_2R_1}{R_1+R_2}$$

This is most easily seen by using the superposition principle, i.e. by setting $V_2=0$ and computing the contribution of $V_1$ and then doing the same for $V_2$. For $I_{Th}$ we get $$I_{Th} = \frac{V_1}{R_1} + \frac{V_2}{R_2}$$ $R_{Th}$ is then given by $$R_{Th} = \frac{V_{Th}}{I_{Th}} = \frac{R_1R_2}{R_1+R_2}$$ The total internal resistance $R_{Th}$ can also be directly obtained by noting that it is the parallel resistance of $R_1$ and $R_2$.

• thank you very much for your valuable comment, it is really helpful and very well explained . but is there any general formula that i can conclude in case of having more sources ? – sahar Apr 13 '13 at 15:40
• As far as I can see there is no general formula. You always have to compute the open-circuit voltage and the short-circuit current, and obviously they always depend on the topology of your circuit. – Matt L. Apr 13 '13 at 18:02
• that's right . Now let's say if i want to measure Thevenin equivalent resistance of the same schematic you just explained, in real time where i can not get the short circuit current (because we are not able short circuit the sources when simulation is running ) or the open circuit voltage (for same reason, we can not make an open circuit when the simulation is running) , then what's the best way to get the equivalent resistance ? – sahar Apr 14 '13 at 12:11
• In your simulation you can probably measure voltages and currents, right? So you just measure the voltage between the terminals indicated in the schematic. Then you just measure the current between the same terminals. If you meters are (close to) ideal, then what you measure is actually the open-circuit voltage and the short-circuit current, respectively. – Matt L. Apr 14 '13 at 13:00
• that's right . but the problem is: if i connect the ammeter to measure the current , then no current will flow to the load (because the resistance through this branch is lower than the branch containing the load) and it will disturb the circuit. that is why i can't actually measure the short circuit current when load is connected to the terminals. so if can't measure the short circuit current i won't be able to measure the Thevenin resistance . – sahar Apr 14 '13 at 13:43