Equivalent resistance of a network

Determine the equivalent resistance across the terminals $$\a-b\$$.

I placed a $$\1V\$$ battery across $$\a-b\$$ and then used mesh analysis on $$\4\$$ loops which resulted in using $$\4\$$ equations. But is it possible to do it in any other way which doesn't need this much equations or complexity? Since the calculators can solve simultaneous equations involving at most $$\4\$$ variables,so I wanted to know the technique where more complexities can be added in the circuit.

• research wye delta transformation Commented Apr 3 at 17:45
• @jsotola I am aware of that transformation. Could you please share how you will be applying that here?Especially whats bugging me in using this is the extra wire of $60$ ohm going up. Thanks in advance. Commented Apr 3 at 18:28
• move the junction at the top right node towards the b node until it is on the left side of the 60 Ω resistor Commented Apr 3 at 18:36
• Please label each resistor with a unique reference like R1, R2 etc... Commented Apr 3 at 20:39

There's a few ways to go when limited by your calculator.

1. Redraw the schematic and examine it for simplifications. Reduce the number of equations needed. Solve using the solver if the remaining equations allow it when using the calculator.
2. Use Cramer's Rule, by hand, if you still have too many equations. (Learn how to use it and practice using it.)
3. Systematically apply Thevenin equivalents, and some logic, and potentially avoid requiring simultaneous equations, entirely.

redraw

simulate this circuit – Schematic created using CircuitLab

(I've grounded b.)

Note that $$\R_1\$$ is in parallel to the other structure. So you can set it aside and focus on the H-structure to the right of it. That's already one simplification and you can see that there are now only two unknown voltages, $$\v_x\$$ and $$\v_y\$$, and this means just two equations:

\begin{align*} \frac{v_x}{R_2}+\frac{v_x}{R_3}+\frac{v_x}{R_6}&=\frac{v_a}{R_3}+\frac{v_y}{R_6} \\\\ \frac{v_y}{R_4}+\frac{v_y}{R_5}+\frac{v_y}{R_6}&=\frac{v_a}{R_4}+\frac{v_x}{R_6} \end{align*}

So solve that using your calculator. Once the voltages at the two unknown nodes are known, you can set the voltage $$\v_a=1\:\text{V}\$$ (or $$\v_a=9\:\text{V}\$$), work out the currents in $$\R_3\$$, $$\R_4\$$, and $$\R_1\$$ and sum them. Then divide that into $$\v_a\$$ to get the total resistance.

Here's the result from a solver using $$\v_a=9\:\text{V}\$$:

solve([Eq(vx/60+vx/30+vx/20,9/30+vy/20),Eq(vy/60+vy/30+vy/20,9/60+vx/20)],[vx,vy])
{vx: 5, vy: 4}


We will be independently finding and using those values below.

Cramer's Rule

I won't get into the details of it. Just look it up. It's not hard.

But the matrix would, in this case, just be just the same thing for those two unknown nodes:

\begin{align*} \left[\begin{matrix} \frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_6}&-\frac{1}{R_6} \\\\ -\frac{1}{R_6}&\frac{1}{R_4}+\frac{1}{R_5}+\frac{1}{R_6} \end{matrix}\right] \left[\begin{matrix}v_x\vphantom{\frac{1}{R_2}}\\\\v_y\vphantom{\frac{1}{R_2}}\end{matrix}\right]&=\left[\begin{matrix}\frac{v_a}{R_3}\\\\\frac{v_a}{R_4}\end{matrix}\right] \end{align*}

The determinant in such a 2x2 matrix is trivial and you can very easily apply Cramer's Rule here.

Apply Thevenin equivalents and logic

One simplification you can make is to Thevenize the left side and the right side.

Suppose you apply $$\v_a=9\:\text{V}\$$. (Yeah, I am picking it.) Then on the left side of the H you would find a Thevenin voltage of $$\6\:\text{V}\$$ and on the right side of the H you would find a Thevenin voltage of $$\3\:\text{V}\$$. (Now you see why I selected that value for $$\v_a\$$.) The Thevenin impedance of each is also the same, $$\20\:\Omega\$$.

So now you can work out the current as the difference of those two Thevenin voltages, $$\3\:\text{V}\$$, divided by the total series resistance, $$\60\:\Omega\$$. That would be $$\\frac1{20}\:\text{A}\$$.

This creates a voltage drop across each resulting series resistor, the two Thevenin equivalent resisors and $$\R_6\$$, of exactly $$\1\:\text{V}\$$, each.

So $$\V_x=6\:\text{V}-1\:\text{V}=5\:\text{V}\$$ and $$\v_y=3\:\text{V}+1\:\text{V}=4\:\text{V}\$$. Trivial.

The three currents that the $$\9\:\text{V}\$$ must supply are $$\\frac{9\:\text{V}}{R_1=20\:\Omega}+\frac{9\:\text{V}-5\:\text{V}}{R_3=30\:\Omega}+\frac{9\:\text{V}-4\:\text{V}}{R_4=60\:\Omega}=\frac23\:\text{A}\$$. Divide that into the original $$\9\:\text{V}\$$ and you get $$\\frac{27}2\:\Omega\$$.

Which is the resistance, by the way.

summary

Get creative, learn a few thinking tools, get good at them, and then when you face a problem you will have a large toolbox available. More like an experienced carpenter facing a project can use their depth and breadth of skills to quickly work out the better way to approach the project.

You do not want to be the person who only knows how to use a chainsaw when facing a kitchen cabinet as a project!

(Please do note that nowhere above do I even consider the idea of a $$\\Delta\leftrightarrows\$$ Y conversion method. When you go look at the conversion equations you can almost immediately see why memorizing that mess is something to avoid if at all possible. And it is frankly never really needed. I haven't used it in more than 30 years' time, except when someone else asks me about it. That said, it is another tool. So I won't tell you not to acquire it. Just that I don't bother with it. The other tools I have are always a better choice.)

Final note: Above, I chose to illustrate three different directions. There are more. The left side of the H is symmetrical and opposite to the right side and that symmetry can be exploited to provide a 4th method I didn't include. The methods above work in asymmetric cases and are therefore more powerful. It would only complicate life to add another that has limited application to special cases. That said, there are times where things are excessively complex and you must consider simplifications like symmetries just to get to the end of the problem. So while I didn't include that here, you should be sensitive to spotting them and, where and when you can imagine how, exploiting them.

After an analysis by inspection and with simple calculations it is found that: