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Electric Potential

  • Thread starter spherific
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http://img519.imageshack.us/img519/3416/32ex28.jpg [Broken]

What is the voltage at points a,b,c,d? I don't know how this works with Kirchoff's loop law. Any help?
 
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Do you at least know the voltage at point c? What about point b, the next easiest?

For the rest, you need to find the current in the loop to find the voltage drop across each resistor. To find this current, you must write the sum of the voltages around the loop with the resistors' voltages represented by IR and set it equal to zero.

You must provide effort and show it for people to help you. No one will solve your homework for you.
 
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At point c the voltage should be 0 because it's grounded. I'm assuming it goes counter-clockwise. So would it be 5V at point B?

Adding up the voltages is:

5 + 4I - 15 + 1I = 0
I = 2

So then at point A it would be

4*2 + 5 = 13

At point D it would be

15 - 13 = 2

and then subtract 2 for point c is

2 - 2 = 0

My question is, why is the question 5 + 4I - 15 + I? I thought resistors cause a drop in voltage to make it 5 - 4I - 15 - I. Is there a concept I'm not understanding? Thanks.
 
393
0
At point c the voltage should be 0 because it's grounded. I'm assuming it goes counter-clockwise. So would it be 5V at point B?

Adding up the voltages is:

5 + 4I - 15 + 1I = 0
I = 2

So then at point A it would be

4*2 + 5 = 13

At point D it would be

15 - 13 = 2

and then subtract 2 for point c is

2 - 2 = 0


My question is, why is the question 5 + 4I - 15 + I? I thought resistors cause a drop in voltage to make it 5 - 4I - 15 - I. Is there a concept I'm not understanding? Thanks.
First off, you have a small error here. The potential across the battery is 15. Therefore, the voltage at the plus sign minus the potential at the minus sign equals 15.
[tex]V_{ad} = V_a - V_d[/tex]

[tex]15 = 13 - V_d[/tex]
Solving the above shows the voltage at d to be -2

to go full circle, you'd then add 2 (instead of subtract 2) and it equals zero.
For clarity, I'll show you:

We know that the voltage drop across the resistor is positive at point c and negative at point d due to the current approaching from point c.
[tex]V_{cd} = V_c - V_d[/tex]
[tex]IR = V_c - V_d[/tex]
[tex]2 = 0 - (-2)[/tex]
[tex]2 = 2[/tex]


On to your question, resistors do cause a voltage drop, but you are adding them because of your notation. Notice, if we multiply your entire first equation by negative -1 on the left and the right, you get the same answer for current and the voltage drop is negative like a drop should be. For ease, however, students are taught to add if they approach a + sign and subtract if they approach a - sign.

Here is the "more correct" way to make your equations but it doesn't really matter. I show this only to clear up confusion. Let's say we assume the current will be clockwise because the 15 volts will overpower the 5 volts (and current will come out of the + on the source making it clockwise.) As we approach the 4ohm resistor, there is a voltage drop:
-4i
as we continue to the 5volt battery, the voltage drops further since we are moving away from the positive and toward the negative end of its terminals:
-4i - 5
then we continue to the next resistor with a voltage drop:
-4i - 5 - 1i
then we see the last battery increases the voltage sense we move from its negative terminal to its positive one:
-4i - 5 - 1i + 15 = 0

Again, if we multiply through with -1, we get the same equation you derived earlier using the less error-prone method of simply choosing the sign of the + or - that you approach:
4i +5 + i - 15 = 0

both equations give the same answer.
 
3
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Thank you for taking the time to post an explanation. You cleared up a lot of confusion. Much appreciated!
 

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