Application of Kirchoff's Laws to circuit with multiple loops

Question

I want to find the current through the 20Ohm resistor in the following diagram, but am stuck with four equations and five unknown currents.

I can apply Kirchoff's current law to each of the nodes either side of 20Ohm resistor, which gives me two equations. Applying Kirchoff's voltage law to the left and right loops yields another two equations. However, I need another relation in order to be able to solve them simultaneously.

Can I assume two "Virtual Earths", so that the bottom rails are both at 0V? That would allow me to make another large loop around the entire circuit to apply Kirchoff's Voltage Law again. However, if this works, I'm not sure why it's a valid approach.

Does anybody have some ideas about how to solve this? Thanks!

Answer 1

I have redrawn the circuit so I can name the nodes and give each component a name. I have also chosen the bottom of Bat1 as our reference 0V so I can assign voltages to each node.

^{simulate this circuit – Schematic created using CircuitLab}

Kirchoff's current law states that the net current into (or out) of any node is zero.

Now because the current in R1 flows out of the positive terminal of BAT1 and the current in R2 flows into the negative terminal BAT2 then \$I_{R1} = I_{R2}\$. You can't have two different currents in a series circuit. This means the current in R3 must be zero.

We can no work out the Voltages at each node:

\$V_{N1}=6 \text{ V } \$, \$V_{N2} = 6 \text{ V } \cdot \dfrac{2 \Omega}{1 \Omega + 2 \Omega} = 4 \text{ V }\$

There is no current in R3 so \$ V_{N3} = V_{N2} = 4 \text{ V }\$

We can see from the loop involving BAT2, R4 and R5 that R4 and R5 each have 5V across them so:

\$V_{N4} = 4 \text{ V } + 5 \text{ V } = 9 \text{ V } \$ and \$V_{N5} = 4 \text{ V } - 5 \text{ V } = -1 \text{ V } \$

Answer 2

Nothing in the diagram indicates that the low sides of batteries are tied to the same potential. You should ask your professor whether or not you can assume that they are.

I suspect that this is a trick question, so I'll assume that the batteries are not tied to the same rail.

Intuition without Kirchhoff. The first thing that catches attention is that these are 2 separate sub-circuits that are connected at only one point with the 20Ω resistor. If I assume that there is current through 20Ω, then it would have to have a return path. That is, another connection between the 2 sub-circuits. But there isn't another connection. So, there should be no current through 20Ω, which makes the problem a lot simpler.

Now add Kirchhoff. How to back-up the intuition with formal Kirchhoff? You wanted to replace the open circuit (∞ Ω) at the bottom of the schematic with a wire (0 Ω) in order to get enough equations. Going from ∞ to 0 just like that... doesn't sound quite right. Instead, draw a resistor with infinite resistance there.
Include this resistance into your new equation. Of course, you'll have I = V/R = V/∞ = 0 in some equations, and that's the same as saying that no current flows through an open circuit.

So there. @Groki, you do the algebra.

Answer 3

Define the currents entering and exiting the nodes N1 and N2.

The current entering the N1 node through the 1 ohm resistor is i1. The current exiting the N1 node through the 2 ohm resistor (on the left) is i2. The current exiting the N1 node and going into the 20 ohm resistor is ix.

KCL tells us: ix + i2 - i1 = 0

i1 must equal i2. Since no other branches are connected to this path.

Simplify, ix = 0 [A]

If you apply the same method for loop 2, you will see the current leaving the N2 node and going into the 20 ohm resistor is also 0 [A]

So the net current going through the 20 ohm resistor is 0 [A]