# Solving Kirchoff's Laws: Are My Answers Correct?

• NikitaMazepin
In summary, the current through R1 I1, I4 through R4, I3 through R3, I2 through R2, and I5 through the 3ohm resistor is 2.375A. Kirchhoff's Current Law states that the currents flowing into any circuit node sum to zero, so the current through R1 I1, I4 through R4, I3 through R3, I2 through R2, and I5 through the 3ohm resistor is zero. Kirchhoff's Voltage Law states that the voltage drops around any circuit loop sum to zero, so the voltage through R1 I1, I4 through R4, I3 through R3, I2 through R2, and If

#### NikitaMazepin

Homework Statement
I am struggling!
Relevant Equations
V = IR
Does anyone know how to do part (a). I labelled the current through R1 I1, I4 through R4, I3 through R3, I2 through R2, and I5 through the 3ohm resistor.

I used Kirchoff’s Laws to make the equations I1 + I4 = 8, 2I1 + 32I3 - 14I4 = 30, and I1 + I3 - I4 = 0.

I solved these equations by subbing in and rearranging to get I1 = 2.375A, I3 = 3.25A, and I4 = 5.625A. Are these correct so far?

Thank you :) Last edited:
I used Kirchoff’s Laws to make the equations I1 + I4 = 8, 2I1 + 3I3 - 14I4 = 30, and I1 + I3 - I4 = 0.
First, your method of naming and labeling all of the currents in the schematic is excellent. Even better, the names make sense, i.e. I2 flows through R2.

Second, I think your equations are a bit sloppy. For example, did you mean ## 3I3 ## or ## 32I3 ##? Also ## I1+I3-I4=0 ## isn't obvious to me (but it may be correct, I'm too busy at the moment to work that out). It isn't one of the most basic KCL equations since these currents don't share a node. You're not adding currents around a loop, are you?

Anyway try again a bit more carefully. At least deal with the 3 vs. 32 question. Show your work and we'll work through this.

Also, just for completeness, let's add some more variable definitions:
## R5 \equiv 3 \Omega ##
## I_5 \equiv ## Current through R5
## I_0 \equiv ## Current through the batteries

You may not need these definitions, but then we'll all be speaking the same language.

• NikitaMazepin and berkeman
First, your method of naming and labeling all of the currents in the schematic is excellent. Even better, the names make sense, i.e. I2 flows through R2.

Second, I think your equations are a bit sloppy. For example, did you mean ## 3I3 ## or ## 32I3 ##? Also ## I1+I3-I4=0 ## isn't obvious to me (but it may be correct, I'm too busy at the moment to work that out). It isn't one of the most basic KCL equations since these currents don't share a node. You're not adding currents around a loop, are you?

Anyway try again a bit more carefully. At least deal with the 3 vs. 32 question. Show your work and we'll work through this.
Sorry I meant 32I3, I'll edit it in the original post.

To be honest I first tried redrawing the circuit with resistors in series and parallel but it got too complicated with too many unknowns.

I thought I2 + I3 - I4 = 0 by imagining the top part of the circuit on its own as a closed loop. This is probably wrong.

I'm not sure where to go next. I have thought of some other equations which may help:
I2 + I5 = 8
I2 + I3 = I1
I5 + I3 + I4
14I4 + 3I5 = 0
2I1 + 11I2 = 0

I thought I2 + I3 - I4 = 0 by imagining the top part of the circuit on its own as a closed loop. This is probably wrong.
Kirchhoff's laws are broken down into two groups:
Kirchhoff's Current Law (KCL): The currents flowing into any circuit node sum to zero.
Kirchhoff's Voltage Law (KVL): The voltage drops around any circuit loop sum to zero.

So, don't mix up currents and loops or voltages and nodes, these won't give useful equations.

Then, at each circuit node you can generate an equation with KCL, like ## I_2 + I_5 = I_0 = 8 A ##. Since this network has 4 nodes* you'll get 4 equations this way. Notice that each term in these equations is a current. If I saw ## I_6 + I_8 +R7I_7 =0 ##, I would immediately know it was wrong because ## R7I_7 ## is a voltage and it doesn't make much sense to add a voltage to a current.

For each circuit loop you can generate an equation with KVL, like ## 2I_1 + 32I_3 - 14I_4 = 0 ##. Since this network has 3 (independent, unique) loops you'll get 3 more equations this way.

For simple components, like resistors, we often use the shorthand of naming the voltage across a resistor as the resistance times the current, like ## 14I_4 ##. But this is skipping the more rigorous approach of naming the voltage (potential) at each node. This would look like ## (V_A - V_B) + (V_B - V_C) +(V_C - V_D) = 0 ## (KVL) and ## (V_A -V_B) = I_{17}R17 ## (Ohm's Law), for example. People rarely do this verbose approach, but it's important to know the steps you may choose to skip. Likewise, there may be a very simple circuit node equation like ## I_3 + I_4 = I_5 ##, which will make some choose to just immediately substitute ## (I_3 + I_4) ## for the label ## I_5 ##. Again this is skipping steps, which is fine if you know that's what you're doing. However, if you get confused, go back and try it again with more basic steps.

So, having labeled all of the variables in your network, you can write down a bunch of equations in a rather mechanical process. Often the more you think about it the greater the chance that you will make a simple mistake, usually a sign error.

This process will actually generate more equations than you actually need. some are redundant and can be ignored.

* I've ignored the node that might (or might not) be between the batteries. Nodes with only two connections are usually ignored, since the equations that result are too simple to be very meaningful, like ## I_8 = I_9 ##. Also note that the voltage across all of the batteries is 54V, not the 30V you used initially.

Last edited by a moderator:

• NikitaMazepin
To be honest I first tried redrawing the circuit with resistors in series and parallel but it got too complicated with too many unknowns.

Time to learn about delta-Y transforms.

Time to learn about delta-Y transforms.
Later. Master KVL and KCL first.

Later. Master KVL and KCL first.