# Calculate using Thevenin's theorem [closed]

How can i get the Vth and Rth on R6 using Thevenin's theorem in this circuit? I have no idea how to do this.

• Why don't you have an idea?
– pipe
Commented Oct 3, 2016 at 14:20
• I remember when I was learning this stuff, there was no Internet. and yes, it was difficult to use only the textbook sometimes to understand and find examples. But today? Come one, you just google "Thevenin examples" or similar, and you get TONS of helpful information, explained at all levels. I see no excuse for such a questions. Commented Oct 3, 2016 at 14:25
• To get the Thevenin voltage assume R6 is open circuit and calculate the voltage across R6. For the resistance replace V1 and V2 with short circuits and calculate the resistance you would measure across R6. Where are you getting stuck? Show us what you have tried so far. Commented Oct 3, 2016 at 14:28
• @WarrenHill I'm stuck at calculation, I don't know which methods to use (Wye-Delta doesn't help) and for the resistance, I can't see which is parallel and which is series. Commented Oct 3, 2016 at 14:34
• If in doubt go back to basics. Remove R6 and imagine I1 comes out of the top of V1 goes through R1, R3 and R4 and back in the bottom of V1. Similarly let I2 flow out of the top of V2 and flows through R2, R3, R5 and back into the bottom of V2. You should be able two simultaneous equations from there using Kirchoff's voltage law solve these for I1 and I2. From this you can work out the volt drop across R4 and R5 and you have your thevenin voltage... Commented Oct 3, 2016 at 14:54

Warren Hill is telling you exactly what the Thevenin Theorem says: The Thevenin voltage is what you'd see across the terminals if you removed $R_6$ and the Thevenin resistance is the Thevenin voltage divided by the current you'd get through the wire if you replaced $R_6$ with a wire (short.) In short, disconnect $R_6$ and measure the voltage with a voltmeter and then measure the current with an ammeter, and divide the voltage measured by the current measured to get the resistance equivalent.

On paper, you have some techniques you are learning. You can replace a pair of series resistors (with nothing else connected between them except themselves) by a single resistor which is the sum of the two resistor values. You can replace a pair of parallel resistors by a single resistor representing the parallel value whose conductance is equal to the sum as the two parallel conductances. You have the ability to switch between a voltage and a series resistor and its Norton equivalent of a current source and a parallel resistor. Between all of these, you can often figure things out.

In your case, that won't be enough. But there is also a superposition law, which can help. So you could apply it and solve the simultaneous equations involved.

But in the final analysis, nodal analysis is probably the most powerful and more consistently applied system. It depends upon the superposition law, of course. But it's easy to learn and use in order to set up the simultaneous equations. You will need to figure out how to solve such equations by hand, if necessary, though. (There is Cramer's rule, which allows you to work out the values one at a time using a formulaic approach.) But there is no escaping that problem if you are aren't allowed a matrix solver on a computer.

If you haven't learned nodal analysis yet, you really should focus on learning it now.

One more thing. You are allowed to establish any one node as $0\:\textrm{V}$ if that helps you simplify the math equations. Doing so won't change any of the resulting analysis. In this case, it would make things overly complex to pick any node other than either side of $R_6$. Here, I'd pick the right side of $R_6$ as my $0\:\textrm{V}$ node and perhaps redraw the circuit to see if that helped me think it through more easily. Something like this:

simulate this circuit – Schematic created using CircuitLab

As you can see, establishing a ground does help slightly. It lets you establish a pair of node voltages, right away, leaving slightly fewer to worry over. The rest is easy to formulate using nodal analysis:

\begin{align*} 1)\:\:\:\:\frac{V_a}{R_1}+\frac{V_a}{R_2}+\frac{V_a}{R_3}&=\frac{V_c}{R_1}+\frac{+24\:\textrm{V}}{R_2}+\frac{V_c}{R_3} \\ 2)\:\:\:\:\frac{V_b}{R_3}+\frac{V_b}{R_4}+\frac{V_b}{R_5}&=\frac{V_a}{R_3}+\frac{V_c-12\:\textrm{V}}{R_4}+\frac{0\:\textrm{V}}{R_5} \\ 3)\:\:\:\frac{V_c}{R_1}+\frac{V_c-12\:\textrm{V}}{R_4}&=\frac{V_a}{R_1}+\frac{V_b}{R_4} \end{align*}

That gives you three equations and three unknowns. Solving this, you know that $V_{th}=V_c-12\:\textrm{V}$.

Now, short the load and redraw the schematic:

simulate this circuit

The current you are interested in is the sum of the two currents shown above. These are the two currents that would flow through your short. To be sure you understand, go back and look at the first schematic above and look at where currents might possibly come from if you were to short out your load resistance. Only two sources of current can arrive there. Hopefully, you can see that fact.

So this one should be easier for you to solve. $R_4$ and $R_5$ are in parallel, and that resulting resistance is in series with $R_3$. The value of $V_d$ can now be easily solved.

$$\frac{V_d}{R_1}+\frac{V_d}{R_2}+\frac{V_d}{R_3+R_4\vert\vert R_5}=\frac{+12\:\textrm{V}}{R_1}+\frac{+24\:\textrm{V}}{R_2}+\frac{0\:\textrm{V}}{R_3+R_4\vert\vert R_5}$$

From that, the value of $V_e$ is a trivial voltage divider computation, as well.

$$V_e=V_d\frac{R_4\vert\vert R_5}{R_3+R_4\vert\vert R_5}$$

Knowing all these, it should be no difficulty at all to figure out the two currents you need to sum up.

Now, with both the Thevenin voltage and currents in hand, you have everything you need for a trivial calculation. The result should be the correct answer.