# Circuit Analysis

1. Oct 13, 2012

### planauts

1. The problem statement, all variables and given/known data

Image: http://puu.sh/1ejYb [Broken]
http://puu.sh/1ejYb [Broken]

2. Relevant equations
Req of n resistors in series = R1 + R2 + ... + Rn
Req of n resistors in parallel = 1/R1 + 1/R2 + ... + 1/Rn

3. The attempt at a solution

The answer is supposed to be 10 ohms. However, I don't know how they got that.

What I was thinking that some of the resistors short circuit and the current does not pass through them. So, it would be like: 3 ohms, then 6 ohms and finally 5 ohms. However, that adds up to 14 ohms and also I'm very sure that's wrong.

Thanks a lot!

Last edited by a moderator: May 6, 2017
2. Oct 13, 2012

### Saitama

It is clear that the 12 ohm resistor short circuit. Now try to see whether the 6 ohm resistor on the right in the upper line and the 6 ohm resistor in the bottom line are in series or parallel.

3. Oct 13, 2012

### PhysicoRaj

The 6Ω resistor in the right at top can be ruled out i think.. bcuz even that is shorted by a wire from the junction of 5Ω and 6Ω resistors at the bottom.

4. Oct 13, 2012

### CWatters

PhysicoRaj - I don't think that's correct.

After removing the 12 Ohm..

The 6 Ohm at the top right is in parallel with the 6 Ohm at the bottom. So delete the 6 Ohm at top right and change the 6 Ohm at the bottom to 3 Ohm.

Then the new 3 Ohm just created is in parallel with the remaining 6 Ohm making 2 Ohms.

At this point you have 3, 2 and 5 in series.

5. Oct 13, 2012

### Staff: Mentor

This is a creative problem. Under examination pressures, where it is often difficult to get ones thoughts clear, there may not be a high score rate on this question.

6. Oct 13, 2012

### hoangkyem

12Ω resistor is short cirrcuit. three 6Ω are parallel.
So, equivalent resistor of circuit is 3+6/3+5=10Ω