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I'm solving this circuit using Ohm's law and Kirchhoff's laws:

circuit diagram

I solved the circuit correctly the second time.

Equation A: $$I_1-I_0-6=0$$

Equation B: $$-120\ \mathrm{V}+I_0\cdot 10+I_1\cdot 50 = 0$$ and when solved, the system of equations comes out to I0 = -3 A and I1 = 3 A.

Once I understood the right way to solve the problem, the setup and math was fine to me. However, the first time I tried to solve the problem, I attempted: $$120\ \mathrm{V}=I_0\cdot 10$$ because using Ohm's law in this way made sense to me. Clearly though, its wrong, and I'm not sure what I'm misunderstanding when it comes to using these laws. My thinking is that Ohm's law directly applied wouldn't work because of the returning current of 6 A, and/or because of the current split at the first node. But I can't reconcile the second one in my head, since the split occurs after the circuit element and (in my understanding) wouldn't affect that resistor. What am I missing? How was I wrong?

I want to apply these equations surely and correctly, knowing why, and not just because I know its what I should do. It's not the setup of the correct formula that eggs me, I'm just missing some very basic understanding of how circuits work.

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  • \$\begingroup\$ You can not do that, because the 10 ohms resistor is not directly connected across 120V. The Ohm's Law says that the voltage across a resistor is equal to the product of the resistance and the current through the resistor electronics.stackexchange.com/questions/511363/… \$\endgroup\$
    – G36
    Feb 13 at 19:31

2 Answers 2

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Your "wrong" equation \$120V = I_0\times 10\$ is wrong because:

  • Ohm's law relates the current through a resistance R to the voltage across that same resistance R. Perhaps what you've done there is try to relate current through a resistor with voltage across a completely different element, the voltage source.

  • If this was an attempt at applying KVL, then it's incorrect. KVL is all about travelling around a loop of things with potential differences across them, "accumulating" any and all changes in potential that you encounter as you go. However, the loop must be complete, and you must end up back where you started. You always ending up at the same potential you started with, for a net change of zero volts. It's like walking in a big loop around town, wherein you may climb hills, and walk down steps, but when you get back to where you started, you must always end up at the same elevation as when you left.

I'll redraw the circuit with some extra annotations, to aid my explanations:

schematic

simulate this circuit – Schematic created using CircuitLab

If your equation \$120V = I_0\times 10\$ is an application Ohm's law, it's wrong because \$I_0\times 10\$ is the potential difference across resistor R1, not voltage source V1. The correct equation would be:

$$ V_{R1} = I_0 \times R_1 $$

If that wasn't the mistake you made, then the equation must be an attempt at applying KVL, which you have misunderstood. Let me rewrite your incorrect equation, to put all non-zero terms on one side:

$$ \overbrace{120V}^{volts} - \overbrace{I_0 \cdot 10}^{volts} = \overbrace{0V}^{volts} $$

That equation seems to be KVL, since all terms, both left and right of the "=", have unit "Volts". Indeed, all terms in such "sums of terms" must all have the same units anyway, because as you've been told a million times, you can't add or compare apples and oranges.

However, it is stating that if you journey from node to node, starting at A, jumping over the source V1 to B, and then across resistance R1 to C, you will observe a total change of potential of 0V. That may (coincidentally) be true in some circumstances (it's not true here), but it's not KVL.

For it to be KVL you must travel around a complete loop, arriving back where you started. You went from A to C, but to apply KVL you must go from A to A. From A, jumping over V1, you encounter a rise of 120V. Jumping over R1, you encounter a fall in potential, and finally, jumping across R2 you will encounter another fall. You're now back at A, for a total change of zero volts:

$$ +120V - V_{R1} - V_{R2} = 0V $$

Of course, now you can substitute voltage terms with expressions resulting from Ohm's law:

$$ +120V - I_0R_1 - I_2R_2 = 0V $$

That is the complete and correct application of KVL.

As a final note, to help hammer the idea home, take a look at what happens if \$R_2 = 0\Omega\$:

$$ \begin{aligned} +120V - I_0R_1 - I_2\times 0 &= 0V \\ \\ +120V - I_0R_1 &= 0V \\ \\ +120V &= I_0R_1 \end{aligned} $$

Now that's the erroneous equation you started with. This would be true if \$R_2 = 0\Omega\$, or in other words, if R1 were connected directly in parallel with source V1.

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You need to find the voltage across the 10 ohm resistor before applying Ohm's law; it's not 120V. You also need to know this voltage for part b) of the problem, to determine the power produced by the current source.

Let's call the voltage at the top node V1. Then we have I0 = (120 - V1)/10, and I1 = V1/50. Substitute these into your equation A. Solving for V1 you will get V1 = 150V.

Applying Ohm's law yields I0 = (120 - 150)/10 = -3A and I1 = 150/50 = 3A

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