Thevenin Method

Question

I need help solving the following circuit using the Thevenin method:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The unknown in this problem is current i5, the correct answer is -1A.

I have already solved for the R(thevenin) by getting rid of all current and voltage sources and replacing them with either an open circuit or a short circuit, the resistance was correctly found to be 4 ohms.

However, I'm not quite sure how to calculate the V(thevenin)in order to reduce this circuit down to a Thevenin circuit. I have tried nodal analysis using node 1 and replacing the currents with differences in potential divided by the resistance but had no luck... any ideas on how to calculate Vth.

Thank you.

My mathematical approach so far:

Once all the dependent sources have been removed and replaced correctly, I see that (((R3+R4) || R6)||R2), calculating this gives ((20||10)||10) --> 4 ohms.

At node 1, I speculated that a current i1 might be going through the R4, the current source is giving a current of i2 = 1A and i3 is leaving the node through the remaining side, so i1+i2 = i3.

I'm providing the first part of an answer below. Is this sufficient? — jonk, Mar 22 '19 at 06:30
@jonk Is there any name for this method of analysis (the one with the vertically laid out circuit)? I want to look into it... — Siddharth Murpani, Mar 22 '19 at 06:35
See the rules I discuss [here](https://electronics.stackexchange.com/a/308882/38098). As far as the idea of being able to assign \$0\:\text{V}\$ to any given (but only one given) node, that's more obvious. A schematic without a ground "floats" so to speak -- there are relative voltages, but no absolute voltage to any node. You can pick any one of nodes and give it an absolute value. This will tether (tie) all the other relative voltages to that value. Using the value ***0*** is a convenience. — jonk, Mar 22 '19 at 06:41
As far as the equation development goes, the left side of the equation is ***out-flowing*** currents and the right side of the equation is ***in-flowing*** currents. Thinking this way helps keep the signs correct and avoids mistakes when writing out the equations. — jonk, Mar 22 '19 at 06:46
Very well, thank you very much! Sorry, not enough rep points to +1 ^^ — Siddharth Murpani, Mar 22 '19 at 06:50
Hehe. I'll add 5 to you. Just to get you started. And you can get another 2 (I think) when you select an answer. — jonk, Mar 22 '19 at 06:51

score 1 · Answer 1 · answered Mar 22 '19 at 06:29

You get to assign \$0\:\text{V}\$ to exactly one node in a circuit. I've chosen one in the following case:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Here, you can see there are only two nodal equations required by the circuit, so that means two equations and two unknowns. Once you know the value for \$V_\text{A}\$ you will know the exact value for \$I_5\$, as indicated in the diagram.

Thus:

$$\begin{align*} \frac{V_\text{A}}{R_1}+\frac{V_\text{A}}{R_2}+\frac{V_\text{A}}{R_3}+\frac{V_\text{A}}{R_4}&=\frac{10\:\text{V}}{R_1}+\frac{20\:\text{V}}{R_2}+\frac{0\:\text{V}}{R_3}+\frac{V_\text{B}}{R_4}\\\\ \frac{V_\text{B}}{R_4}+\frac{V_\text{B}}{R_5}&=\frac{V_\text{A}}{R_4}+\frac{0\:\text{V}}{R_5}+1\:\text{A} \end{align*}$$

Are you able to go from here?

(You could also just Thevenize your way through this without nodal analysis, I suppose.)

Thevenin Method

1 Answers1