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

circuit

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!

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  • \$\begingroup\$ There are only two loops, and they are independent of each other. You can't assume that both bottom rails are at the same potential. The 20 ohm resistor can't carry any current, because it wouldn't have a return path. \$\endgroup\$ – apalopohapa Nov 7 '14 at 19:29
  • \$\begingroup\$ What are your five unknown currents? I count 3 -- one in each loop, and the current through the \$20\Omega\$ resistor. \$\endgroup\$ – Null Nov 7 '14 at 19:30
  • \$\begingroup\$ Hi. Thanks for the comments. I think the current through each resistor is an unknown, unless you assume the current through the 20ohm resistor is zero, which is what the question is asking for. Can't the batteries sink current, nullifying the need for a return path? \$\endgroup\$ – groki Nov 7 '14 at 20:15
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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.

schematic

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 } \$

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  • \$\begingroup\$ Does the reasoning of the current out of the positive terminal being equal to the current into the negative terminal not also hold for any arbitrary value of the current through R3? Let the current through R3 be I3, then as long as it is split equally through R2 and R1, the current in/out of the battery will still be the same? \$\endgroup\$ – groki Nov 7 '14 at 21:47
  • \$\begingroup\$ No the current \$I_{R1}\$ flows out of the the positive terminal of Bat1 and flows through R1 left to right as drawn. The current \$I_{R2}\$ flows through R2 top to bottom as drawn then into the bottom of BAT1. These both have to be equal. Now assume we have a current \$I_{R3}\$ flowing in R3 left to right as drawn drawn then since Kirchoff states that the net current into a node must be zero. \$I_{R1} - I_{R2} - I_{R3} = 0\$ given \$I_{R1} = I_{R2}\$ then the only possible conclusion is \$I_{R3} = 0\$. \$\endgroup\$ – Warren Hill Nov 7 '14 at 22:04
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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.

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  • \$\begingroup\$ Hi. Thanks for your answer. I'm not sure I understand the intuition: isn't it possible for one of the batteries to sink current? Also, why does adding in an infinite resistor help with the formal Kirchoff reasoning? It just tells you that the current through the "vertical" 2ohm resistors is equal to the current through them - a valueless tautology. \$\endgroup\$ – groki Nov 7 '14 at 20:12
  • \$\begingroup\$ Also, if you want to do use an infinite resistance to create a massive loop around the circuit for Kirchoff's Voltage Law, that means you'd have to calculate V = IR = infinity * 0, which is unhelpful. \$\endgroup\$ – groki Nov 7 '14 at 20:30
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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]

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