spherific said:
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.
V_{ad} = V_a - V_d
15 = 13 - V_d
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.
V_{cd} = V_c - V_d
IR = V_c - V_d
2 = 0 - (-2)
2 = 2
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.
