Welcome to blog help for PHY 122! Here's how it works.
For each homework question and lab report we will make a post, this will probably contain a few tips on what the problems are about and how to solve them. If you are stuck on something then instead of emailing us directly you should post a comment in reply to the relevant post. We will try to guide you through tough points and help you understand the problems and the concepts behind them.
Thursday, September 23, 2010
Chapter 18, Problems 7-9
These problems require application of Kirchhoff's rules for more complicated DC circuits. See section 18.4
This post came before I broke the chapter into separate streams. Anonymous said... Hi Professor, I'm having a lot of trouble with question number eight. I set up my three equations: 12-(.01)I1-(.9)I2-7.5=0, and then 7.5+(.9)I2-(.06)I3=0, and finally I2+I3=0. Both my loops are in the clockwise directoin. Each time I solve the equation, I can't arrive at the correct answer, and I also have too many things cancel out, often leaving me with a current of zero. Am I assigning the incorrect signs?
I can't completely follow your equations without knowing how your drew I1, I2, and I3. So let me guess that you drew them all pointing upward, thru the live battery, dead battery, and starter, respectively. Also, I have to guess that your component values are R=0.90 Ohm and V = 7.5 Volts (dead battery). Left side loop would be: 12 - 0.01*I1 + 0.90*I2 - 7.5 = 0 You can work up a similar equation fot the right side loop, involving I2 and I3. Your equation I2+I3 can't be right (if I understand how you defined the I's) because that would mean there is no current I1 flowing in the leftmost branch. It is the sum of ALL of the currents flowing into the top node that must be zero.
hi professor, i'm a bit stumped on questions #7. i know that I1 + I2 = I3, but i have no idea how to approach this problem. The presence of the extra unknown battery voltage confuses me even more.
I don't understand problem 9 b. I set up a circuit with one piece of the current flowing from the battery going to one circuit and another piece going to another circuit. I said I1 = I2 + I3 with the I2 current flowing to the headlights and I3 current flowing to the starter motor. I also am using delta V = Emf - I1 (Rbat). Is this correct? I used Kirchhoffs rules after that point to solve for Rstarter.
Using Kirchoff's Rule, we know that I1 + I2 = I3. We also know that the sum of the potential differences across all the elements of a closed circuit "loop" must equal to zero.
For example, if you drew a loop going clockwise in the second half of the circuit, the sum of 24V, voltage of R1, and voltage of R2 would have to equal zero. It should look something like this:
+(-)24V +(-)I1*R1 +(-)I2*R2 = 0
Keep in mind when setting up the equation, depending on the direction of your loop compared to the direction of the current, the signs of the change in potential may either be - or +. PAGE 604 gives a good detailed explanation on when the change is - or +.
Once you have solved that equation, you will have determined the value of I2. Then, using the junction rule (I1 + I2 = I3), you can easily solve for I3.
You have Rint and the current so the key is to first find the voltage across Rint and from there you can find the potential difference across the bulbs by the differences of the voltage across Rint and your total voltage.
Prof. Stephens talked about this near the beginning of today's lecture so the first few slides may help clear the concept up.
For 9b: Robert's answer isn't quite right (although WA may accept it within its 1% numerical tolerance). The problem is that you don't know the total load resistance. You know current through the starter, but not the voltage across it. So this is a Kirchhoff's laws problem. Let the current flowing through the battery be Ibatt, and the current flowing through the headlights be Ihl. You are given that the starter current is 35 Amps (or whatever numerical value you have). So KCL says Ibatt=Ihl+35. Now consider the loop that goes through the battery, the headlights, and the internal resistance of the battery. That loop equation will have three terms - the battery EMF, the headlights, and the internal resistance of the battery. You can solve that for Ihl, and since you know their resistance, that gives you the voltage across the headlights.
8 comments:
This post came before I broke the chapter into separate streams.
Anonymous said...
Hi Professor,
I'm having a lot of trouble with question number eight. I set up my three equations: 12-(.01)I1-(.9)I2-7.5=0, and then 7.5+(.9)I2-(.06)I3=0, and finally I2+I3=0. Both my loops are in the clockwise directoin. Each time I solve the equation, I can't arrive at the correct answer, and I also have too many things cancel out, often leaving me with a current of zero. Am I assigning the incorrect signs?
I can't completely follow your equations without knowing how your drew I1, I2, and I3. So let me guess that you drew them all pointing upward, thru the live battery, dead battery, and starter, respectively. Also, I have to guess that your component values are R=0.90 Ohm and V = 7.5 Volts (dead battery).
Left side loop would be:
12 - 0.01*I1 + 0.90*I2 - 7.5 = 0
You can work up a similar equation fot the right side loop, involving I2 and I3.
Your equation I2+I3 can't be right (if I understand how you defined the I's) because that would mean there is no current I1 flowing in the leftmost branch. It is the sum of ALL of the currents flowing into the top node that must be zero.
hi professor, i'm a bit stumped on questions #7. i know that I1 + I2 = I3, but i have no idea how to approach this problem. The presence of the extra unknown battery voltage confuses me even more.
I don't understand problem 9 b. I set up a circuit with one piece of the current flowing from the battery going to one circuit and another piece going to another circuit. I said I1 = I2 + I3 with the I2 current flowing to the headlights and I3 current flowing to the starter motor. I also am using delta V = Emf - I1 (Rbat). Is this correct? I used Kirchhoffs rules after that point to solve for Rstarter.
Re: Anonymous, #7:
Using Kirchoff's Rule, we know that I1 + I2 = I3. We also know that the sum of the potential differences across all the elements of a closed circuit "loop" must equal to zero.
For example, if you drew a loop going clockwise in the second half of the circuit, the sum of 24V, voltage of R1, and voltage of R2 would have to equal zero. It should look something like this:
+(-)24V +(-)I1*R1 +(-)I2*R2 = 0
Keep in mind when setting up the equation, depending on the direction of your loop compared to the direction of the current, the signs of the change in potential may either be - or +. PAGE 604 gives a good detailed explanation on when the change is - or +.
Once you have solved that equation, you will have determined the value of I2. Then, using the junction rule (I1 + I2 = I3), you can easily solve for I3.
Hope this helps!
Thanks Jen. Everything OK now?
I still don't get number 9 part b. I left a comment but I think it got missed.
For 9b.
You have Rint and the current so the key is to first find the voltage across Rint and from there you can find the potential difference across the bulbs by the differences of the voltage across Rint and your total voltage.
Prof. Stephens talked about this near the beginning of today's lecture so the first few slides may help clear the concept up.
For 9b:
Robert's answer isn't quite right (although WA may accept it within its 1% numerical tolerance). The problem is that you don't know the total load resistance. You know current through the starter, but not the voltage across it.
So this is a Kirchhoff's laws problem. Let the current flowing through the battery be Ibatt, and the current flowing through the headlights be Ihl. You are given that the starter current is 35 Amps (or whatever numerical value you have). So KCL says Ibatt=Ihl+35. Now consider the loop that goes through the battery, the headlights, and the internal resistance of the battery. That loop equation will have three terms - the battery EMF, the headlights, and the internal resistance of the battery. You can solve that for Ihl, and since you know their resistance, that gives you the voltage across the headlights.
Post a Comment