Finding the Current in a circuit

  • #1
John Wiggum
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Homework Statement


The diagram below shows a circuit where; R1 = 5.0 Ω, R2 = 8.0 Ω, R3 = 1.0 Ω, V1 = 16.8 Volts, V2 = 1.5 Volts, and V3 = 39.6 Volts. What is the value of I1? In solving this problem, initially pick the current directions as shown. If the actual current turns out to be in the opposite direction, then your answer will be negative. If you get a negative number, enter it as negative.

Here is a diagram of the circuit with assumed directions of current.\

It should also be noted that my Physics class counts crossing a battery in a KVL loop from its negative terminal to its positive terminal as a positive voltage, and crossing a resistor in a KVL loop in the flow of net positive charge as a negative contribution to voltage for the KVL loop equations.


Homework Equations


[/B]
Kirchoff's Circuit Law: ##Σ I_{in} = Σ I_{out}##

Kirchoff's Voltage Law: ##Σ_{closed~loop} ΔV_i = 0##

Ohm's Law: ##V=I*R##

The Attempt at a Solution


[/B]
So based on the problem picture, and using KCL I determined ##I_1 + I_2 + I_3 = 0## for one of my equations. Then using KVL and drawing two circuit loops shown here I derived two equations. The equation for the 1st loop I found to be ##-I_1*R_1 - V_1 + I_2*R_2 + V_2 = 0##, and I found the equation from the second loop to be ##-I_2*R_2-V_2+I_3*R_3+V_3=0##. When setting up a system of equations and solving them I get ##I_1##=0.5833 amps. This is wrong though and I believe it is a sign error in one of my KVL equations. However, I am unsure of what sign is the issue.
 

Answers and Replies

  • #2
TSny
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Welcome to PF!
So based on the problem picture, and using KCL I determined ##I_1 + I_2 + I_3 = 0## for one of my equations.
This isn't correct. You should be setting up the equation ##Σ I_{in} = Σ I_{out}##. Which junction are you using to set up the equation? For that junction, which currents are "in" and which are "out"? [EDIT: Your equation is correct!]

Then using KVL and drawing two circuit loops shown here I derived two equations. The equation for the 1st loop I found to be ##-I_1*R_1 - V_1 + I_2*R_2 + V_2 = 0##, and I found the equation from the second loop to be ##-I_2*R_2-V_2+I_3*R_3+V_3=0##. When setting up a system of equations and solving them I get ##I_1##=0.5833 amps. This is wrong though and I believe it is a sign error in one of my KVL equations. However, I am unsure of what sign is the issue.
You have some sign errors in both of these equations. Make sure that you start at one point of the loop and go around the loop in one direction (clockwise or counterclockwise) until you return to the starting point. Be sure to follow the sign rules exactly as you stated them below:
It should also be noted that my Physics class counts crossing a battery in a KVL loop from its negative terminal to its positive terminal as a positive voltage, and crossing a resistor in a KVL loop in the flow of net positive charge as a negative contribution to voltage for the KVL loop equations.
 
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  • #3
John Wiggum
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Welcome to PF!
This isn't correct. You should be setting up the equation ##Σ I_{in} = Σ I_{out}##. Which junction are you using to set up the equation? For that junction, which currents are "in" and which are "out"?

You have some sign errors in both of these equations. Make sure that you start at one point of the loop and go around the loop in one direction (clockwise or counterclockwise) until you return to the starting point. Be sure to follow the sign rules exactly as you stated them below:
Thanks for the quick reply! I guess I should have clarified in my original statement I was using the middle top junction of the circuit diagram to define my KCL equation. As for why I decided that the combination of all three of them is 0 is because in the original problem picture it draws all currents leaving the battery at the positive terminals and I thought all currents would then meet in the middle and equal zero. This would mean that at least one of my three currents is negative meaning the original direction of at least one of the currents is wrong, but I thought the sign conventions would take care of that.
 
  • #4
TSny
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Thanks for the quick reply! I guess I should have clarified in my original statement I was using the middle top junction of the circuit diagram to define my KCL equation. As for why I decided that the combination of all three of them is 0 is because in the original problem picture it draws all currents leaving the battery at the positive terminals and I thought all currents would then meet in the middle and equal zero. This would mean that at least one of my three currents is negative meaning the original direction of at least one of the currents is wrong, but I thought the sign conventions would take care of that.
OK. Yes, all currents are going into the top junction. So, you are correct here. I was wrong.

I don't understand your handwritten drawings. In loop 1 you have I2 as going down through R2. But the original drawing has it going up. Also, in this loop you have the current along the bottom of the loop labeled as I2, but shouldn't it be I1?

In loop 2, shouldn't the current along the horizontal top part of the loop be I3 toward the left instead of I2 toward the right? Also, shouldn't I3 be upward through R3? And I3 along the bottom of the loop should be toward the right.
 
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  • #5
John Wiggum
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OK. Yes, all currents are going into the top junction. So, you are correct here. I was wrong.

I don't understand your handwritten drawings. In loop 1 you have I2 as going down through R2. But the original drawing has it going up. Also, in this loop you have the current along the bottom of the loop labeled as I2, but shouldn't it be I1?

In loop 2, shouldn't the current along the horizontal top part of the loop be I3 toward the left instead of I2 toward the right? Also, shouldn't I3 be upward through R3? And I3 along the bottom of the loop should be toward the right.
Oh good catch! My current is incorrect for loop 2 Thank you! When setting up a KVL loop I was told to pick either a clockwise or counter clockwise direction for the current flow, and as long as I have my signs correct for my KVL equation it should still work out.
 
  • #6
TSny
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Oh good catch! My current is incorrect for loop 2 Thank you! When setting up a KVL loop I was told to pick either a clockwise or counter clockwise direction for the current flow, and as long as I have my signs correct for my KVL equation it should still work out.
Yes, that's right. Which way did you go around loop 1? Clockwise or counterclockwise? Do you still think your equation for this loop is correct?
 
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  • #7
John Wiggum
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Yes, that's right. Which way did you go around loop 1? Clockwise or counterclockwise? Do you still think your equation for this loop is correct?
OK so I went dark on Sunday, but I just wanted to let you know that, Yes I got it! It ended up being a sign error with my two resistors on both my loops. In the end I got my final KVL equations to be ##Loop_1, 0 = -I_1*R_1+V_1+I_2*R_2 - V_2 = 0##, and ##Loop_2, 0 = I_2*R_2 - V_2 - I_3*R_3 + V_3 = 0##.

Thanks for all your help! It turns out that many of the help pages I was looking at on the internet had opposite sign conventions to the ones taught in my class, and I was thus mixing them together to produce wrong equations.
 
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