# Find Voltage and Current

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1. Sep 17, 2016

1. The problem statement, all variables and given/known data
Find the voltage v and the currents i1 and i2 for the circuit below.

2. Relevant equations
V=IR

3. The attempt at a solution
I would combine the 10 ohm and 40 ohm, and the 18 ohm and 9 ohm, so that they are in series with the 6 ohm resistance. Then I would use V=IR to get the current, but not sure if this is the correct approach.

2. Sep 17, 2016

### Staff: Mentor

I think you mean you could use that method to find the voltage v across the 5 A current supply. Yes, that would work.

You're still left with finding the individual currents $i_2$ and $i_2$. You know about voltage dividers (from a previous thread), but do you also know about current dividers?

3. Sep 17, 2016

No, I am not aware of current dividers

4. Sep 17, 2016

So I looked it up, and this is what I calculated. I made all resistances in series and added them and got 20 ohms. V=IR = (2A)(20 ohms) = 40V. Then i1 = itotalRtotal / R1 = (2A)(20 ohms) / (12 ohms) = 3.3A

5. Sep 17, 2016

i2 will also be the same as i1 according to my calculations since resistances across each will be equal

6. Sep 17, 2016

I am using different values then the ones show in the picture, they are 2A for the total current instead of 5A, and from left to right, the resistances are 12, 24, 8, 6, and 12

7. Sep 17, 2016

### Staff: Mentor

A current can't be larger than the current source. So your 3.3 A result should send up a flag that something's off.

Your problem is that you've got the current divider equation a bit mixed up. You want to multiply the total current by the resistance of the "other" path divided by the total resistance. So for example, using the figure I posted in post #2,
$$i_1 = I \frac{R_2}{R_1 + R_2}$$
Note that the "other" path in this case is comprised of $R_2$.

8. Sep 17, 2016

Therefore, i1 = (2A)(12 ohms) / (12 ohms + 12ohms) = 1 amp
Since R1 and R2 are equal, i1 and i2 will also have the same value.
Would that be correct?

9. Sep 17, 2016

### epenguin

Nothing wrong with what you said and did – but it's a bit pedestrian and following textbook routine.
You should be able to say the currents off the top of your head!
That is what gneil is getting at saying "current dividers".
Certainly off the top of your head you can say immediately what the current is in the 6 Ω resistor. OK you're asked what it is in the branch parts. Current flows the parallel conductors just in proportion to their conductances.

10. Sep 17, 2016

### Staff: Mentor

No, There are two separate current dividers to consider: one divider consists of a 12 Ω and a 24 Ω resistor while the other consists of a 6 Ω and a 12 Ω resistor. See the image below. One current division is shown in blue, the other green.

Last edited: Sep 17, 2016
11. Sep 18, 2016

### epenguin

Maybe repeating, but you should find this really easy.

Referring for instance to your original #1 values, Just because the arithmetic is particularly simple, for the left-hand branch 5A enters and leaves, dividing itself between conductances which are in the ratio 4:1. the current distributes itself in that ratio.