Electronic – How to simplify this circuit to find resistor voltage

basickirchhoffs-lawsvoltage divider

In my circuits class, I am presented with the following exercise.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The problem is to find the voltage across resistors A and B. The most likely solution (given the section) is to use simplification by analysis. I tried to solve for B first. My attempt included cutting the circuit in two parts (all of the resistors and just B), simplifying the part of the circuit that was not B, and then solving the resulting circuit. This yielded a parallel circuit.

I know that is not supposed to yield a parallel circuit which implies a direct voltage path from the source to resistor B.

The book as has a similar example, but in this one the two ohm resistor is to the right of the 3 ohm resistor, making simplification vastly easier. (I seem to have missed the lecture on this.)

Could someone be so kind as to point me in the right direction?

Best Answer

Your approach, if not your execution, is sound .

If you're familiar with Thevenin equivalent circuits, you know that you can replace the entire circuit to the left of the B resistor with a voltage source in series with a resistor. Once you've done this, voltage division will give you the voltage across B.

For A, replace all the resistors to the right of A with an equivalent resistor which is in parallel with A. Then, once again, apply voltage division to find the voltage across A.

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KCL for this Supernode

For starters, I don't understand how the numerator on these fractions ends up as anything but one....

I've left several comments pointing out careless errors in your equations but this deserves an answer even if the answer is elementary algebra.

Take, for example your first (corrected) equation:

$$\frac{1}{10}v_1 + \frac{1}{20}v_3 +\frac{1}{2}(v_3-v_2)+\frac{1}{2}(v_1-v_4)=0$$

Group the terms involving, for example, \$v_1\$

$$v_1(\frac{1}{10} + \frac{1}{2}) = v_1 \frac{3}{5}$$

Electronic – Learning loops, Kirchhoff’s Laws

I think you may have an issue with one of the signs. Going around the loop, I get:

12 - 4 i - 2 Vo + 4 - 6 i = 0

Looks like you got the sign on the 6i term flipped.

So, if we take Vo = -6 i, you can calculate:

12 - 4 i + 12 i + 4 - 6 i = 0
16 + 2 i = 0
i = -8
Vo = 48

If you're stuck on understanding how to write this equation, the way to remember it is 'what goes up must come down'. Sum up the voltage across each element as you go around the loop. Since you go into the - side of the 12 volt source, you gain 12 volts. Then you pass through a resistor, and you lose 4 i volts. Then you go the wrong way through the dependent source and you lose 2 Vo volts. Then you go through the 4v source and gain 4 volts. Then you go through the final resistor and lose 6 i volts. You're back where you started, so all of that adds up to zero. Then you figure out what Vo is - if the current is flowing in the direction of the arrow, it will create a voltage drop in opposition to how Vo is marked, so Vo = -6 i. After that, finding i is just algebra.

Best Answer

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KCL for this Supernode

Electronic – Learning loops, Kirchhoff’s Laws

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