# Ohms Law Voltage/Potential Difference [duplicate]

This question is an exact duplicate of:

Three 2.0 Ω resistors are connected in series to a 12 V battery. What is the potential difference across each resistor? � This question requires Ohms law because V is needed right? So I would use I = V/R and use Req = r1 + r2 + r3... then does I(Current) = 12?

so 12 = V/6 Im confused.

## marked as duplicate by Brandon Enright, Qmechanic♦Apr 9 '14 at 22:34

This question was marked as an exact duplicate of an existing question.

• $R_\mathrm{eq}$ looks right. How did you calculate the current to arrive at 12 A? (Assuming you do mean I(Current) = 12 Amp.) – garyp Apr 9 '14 at 20:57
• The 12V battery is not the current so how would I calculate it? I = v/r but I dont have v? the correct answer is 4V for the potential resistance. @garyp – Help Apr 9 '14 at 21:00
• It looks like you have a battery providing a voltage, and the resistance across the battery. You've also demonstrated that you know Ohm's Law. Maybe you should draw a sketch with the three separate resistors replaced by the one equivalent resistor. See if it helps to look at that sketch. – garyp Apr 9 '14 at 21:05
• The 12V battery is not the current so how would I calculate it? I = v/r but I dont have v? the correct answer is 4V for the potential resistance. @garyp I mean I got the current to be 2, voltage to be 12, and resistance to be 6, but the correct answer says the potential difference is 4V? Is it 4 because since the V = 12 and it is asking for potential difference across each resister and there are 3, so 12/3 = 4V? lol oh... my bad – Help Apr 9 '14 at 21:06
• Your analysis is correct for this highly symmetric case. All of the resistors have the same value. Suppose the three resistors were $R_1 = 1\;\Omega$, $R_2 = 2\;\Omega$, $R_3 = 3\;\Omega$. Can you calculate the voltage across $R_1$? Doing this exercise provides a method to solve these problems that will work in more general situations than does your method. – garyp Apr 9 '14 at 21:11

If the three resistors are connected in series, the total effective resistance is $3\times 2\Omega=6\Omega$ and the current is $I=U/R_{\rm tot}=12/6{\rm A}=2{\rm A}$, and the potential across each resistor is $IR=2\times 2{\rm V}=4{\rm V}$.
However, an easy way to do it is, since we know in this case the current through each resistor is the same, so the potential $U$ must be divided evenly, which means the potential across each resistor is $12{\rm V}/3=4{\rm V}$.