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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?


|                 |                |               |
--(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?

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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?

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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.

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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.

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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,

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