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What I know is with a current source, the current through the resistor should be the same as the source. But here I have two current sources so would this affect the value? I need to work out current through each of the resistors.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I what I know is with a current source, current thorough the resistor should be the same as the source. but here I have two current sources so would this affect the value? because I need to work out current through each of the resistors. \$\endgroup\$ – user31328 Nov 6 '13 at 1:21
  • \$\begingroup\$ I've just incorporated your comment into the question, you can edit the question yourself at any time and that's better than including additional info in the comments. \$\endgroup\$ – PeterJ Nov 6 '13 at 2:45
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Since someone else has provided an answer, I shall provide a more intuitive approach.

Since the two current sources are in parallel, combine the 2A and 1A sources into one 3A source and then you have the canonical current divider circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

$$I_{R1} = 3A \dfrac{R_2}{R_1 + R_2} = 3A \dfrac{6}{3 + 6} = 2A $$

etc.

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R1 and R2 are in parallel. Via Ohm's law:

3ohm || 6ohm = 2ohm

I1 and I2 are in parallel with regards to the load. KCL:

2A + 1A = 3A

The current sources are in series with the resistors. Ohm's law:

2ohm * 3A = 6V

The same voltage is across each resistor. Ohm's law:

6V / 3ohm = 2A
6V / 6ohm = 1A

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Your current sources are in parallel, so they can be added (I = I1 + I2)

Then, since your resistors are in parallel, they have the same voltage, and the current is divided proportionally between them:

Current Through R1 = R2/(R1+R2) x I
Current Through R2 = R1/(R1+R2) x I

To double-check:

In this case, R1 = 1/2 R1, so the current through R1 will be twice the current through R2.

So you have a special case, where the current sources and resistor values are both in the ratio of 2:1, so the answer can be seen immediately. But for different current source and resistor values, the equations above first calculate the total current, then divide that proportionally between the two resistors.

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  • \$\begingroup\$ This is in error. The larger current is through the lessor resistance. You've used the voltage divider ratio rather than the current divider ratio. \$\endgroup\$ – Alfred Centauri Nov 6 '13 at 11:20
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Parallel resistors have same voltage on them, and therefore their current becomes related their resistance according to ohm law.

V = I x R

Thereby, if the current that flows through R2 is to be I1, then current of R1 would be 2xI1.

Current sources must flow their electrons on resistors eventually. Thus current total of resistors must be equal to total current.

I1 + 2xI1 = 3xI1 = 3A

I1 = 1A

From this, 2A flows through R1, and 1A flows through R2.

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