# Is this superposition example wrong?

Discussion in 'General Electronics Chat' started by hunterage2000, May 2, 2010.

May 2, 2010
474
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2. ### Ghar Active Member

Mar 8, 2010
655
73
The higher resistance will always have less current than the lower resistance when in parallel.
In this case you have R1 = 4 ohms and R2 = 2 ohms.

Since they're in parallel in that 7 V circuit, R2 must have twice as much current as R1.
R3 will get the sum of those two currents.

In the 28 V circuit R2 and R3 are in parallel, and R3 is 1 ohm. Therefore the situation is swapped, with R3 having twice the current of R2.
R1 will get the sum of those two currents.

3. ### hunterage2000 Thread Starter Senior Member

May 2, 2010
474
1
I used for I1 2/2+4 x 3 = 1 and I2 4/2+4 x 3 = 2, cant see where im going wrong

4. ### Ghar Active Member

Mar 8, 2010
655
73
You really need to use brackets.

I1 = 2/(2+4) * 3 = 1

So I'm not sure what the issue is. I1 is the current through R1, yes?
Then it agrees... 4 ohm R1 gets 1 A, 2 ohm R2 gets 2 A.
Where's the disagreement?

5. ### hunterage2000 Thread Starter Senior Member

May 2, 2010
474
1
Just I thought as R1 = 4 R2 is 2 then I1 would be I1 = 4/(2+4) * 3 = 2 not 1 which it says in the tutorial. Is this the correct way of doing this because the 28v circuit was correct.

6. ### Ghar Active Member

Mar 8, 2010
655
73
With current division the numerator is the resistance in the other branch.

So:
I1 = Itotal * R2 / (R1 + R2)
I2 = Itotal * R1 / (R1 + R2)

The 28V circuit can do the same thing:
Itotal = 6 A.
R2 = 2 ohms
R3 = 1 ohm

I2 = 6 * 1 / (2 + 1) = 2 A
I3 = 6 * 2 / (2 + 1) = 4 A

7. ### hunterage2000 Thread Starter Senior Member

May 2, 2010
474
1
cheers mate thanks for clearing that up

8. ### Ghar Active Member

Mar 8, 2010
655
73
You're welcome.