Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I just recently finished a test on Physics, and when receiving my scores back I missed a question. Here's what it was if I can remember it correctly:

Three resistors (10.0, 15.0, and 6.00 ohms) are parallel and the circuit is powered by 12.0V. How much current (in amps) is the 6.00 ohm resistor using?

So:

|-----------------|----------------|---------------|
|                 |                |               |
--(12.0 V)        <(10.0 ohms)     <(15.0 ohms)    <(6.00 ohms)
-                 >                >               >
|                 |                |               |
|                 |                |               |
----------------------------------------------------

So, I assumed that since you can combine parallel resistors into one equivilent resistor, they use the same amount of current.

So I just found the current used (4.00 A), and assumed that the 6.00 ohms resistor used the same current. The teacher, however, stated that the correct answer was 2.00 A.

Am I right or wrong? Why?

share|improve this question
I'm tagging this homework because it's a question of educational value (even though I recognize it's not an actual homework question). – David Zaslavsky May 12 '11 at 17:54
Thank you David, I was debating whether to use it or not. :) – Ramp May 12 '11 at 17:56
A resistor does not "use" current! The current flows through the resistor. – Georg May 13 '11 at 2:21

4 Answers

up vote 1 down vote accepted

Think about this: the current through a certain circuit element represents the amount of charge that flows through that element per second. In a case like this, it's electrons that carry the current, so the amount of current is related to the amount of electrons that flow per second.

Now, you found that a total of $4\ \mathrm{A}$ of current flow through the circuit. Suppose that corresponds to a total of $N$ electrons per second. Are all $N$ of those electrons going to go through the $6\ \Omega$ resistor?

share|improve this answer
Okay, I see what you mean. Then how would I calculate that? I just feel rather cheated or something because he never taught us how to do that. I guess you just make a ratio of the resistors and then take that portion of the total current? How would that come out to 2.00 A, then? – Ramp May 12 '11 at 17:51
@Ramp: do you know what property of parallel resistors has to be the same? (It's not the current) – David Zaslavsky May 12 '11 at 17:53
@David: Are you talking about voltage? – Ramp May 12 '11 at 17:57
@Ramp: yep, that's the one ;-) Knowing that the voltage is the same across parallel resistors (which I would hope you were taught) is the fact you need to figure out how much current flows in each. – David Zaslavsky May 12 '11 at 19:42
I doubt that going down to electrons/sec helps a lot for someone who is thought to apply Kirchoff, not more. – Georg May 13 '11 at 2:23
show 2 more comments

Only if the resistors are the same, they carry the same amount of current. Otherwise, the current is distributed according to the ratio of their resistances.

share|improve this answer

The answer is 2.0 Amps.

This is a parallel circuit. Each resistor sees exactly the same voltage (assuming negligible wire losses). The current through each resistor is 12V/(R Ohms) = X Amps, where X = 12/R. The currents through the three resistors (starting from the left) are 1.2A, 0.8A and 2.0A. TOTAL current is 0.8+1.2+2 = 4.0 Amps.

If you wanted to combine the parallel resistances into one equivalent resistance, you could use the formula Reqv = 1/((1/R1)+(1/R2)+(1/R3)). But in this case there is an easier way: Reqv = 12V/4A = 3 Ohms. We can check this using the first equation: Reqv = 1/(0.1 + 0.066667 + 0.166667) = 1/(0.33333) = 3.00 Ohms.

share|improve this answer

I would suggest investigating Kirchoff's Laws of 'Current' and 'Voltage'.

I find the following website very helpful in teaching some circuit theory and this problem is covered as well, http://www.utwired.engr.utexas.edu/rgd1/lesson03.cfm

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.