# Use resistor simplification to find current across the 10 ohm resistor

1. Jan 23, 2012

### TheRealDoodle

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

Use resistor simplification to find current across the 10ohm resistor

2. Relevant equations

V=ir p=vi

3. The attempt at a solution

Any help you can give me is much appreciated. When I did it i got down to a parallel circuit with a 30ohm and a 100ohm but didnt get a round answer and i thought the teacher said theyre all rigged to come out right.

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2. Jan 23, 2012

### Staff: Mentor

Can you show the work for your attempt?

3. Jan 25, 2012

### TheRealDoodle

Yes, here is my work, sorry for not posting originally. I hadnt read the rules yet.

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4. Jan 25, 2012

### Curious3141

I get the same answer you do. The quicker way to do it (after you disregard everything distal to the short) is to observe you're left with 4 parallel resistances 40||25||25||200, the effective resistance of which is more quickly worked out with 1/R = (1/R1 + 1/R2 + 1/R3 +1/R4). Then add that to the 10Ω in series.

5. Jan 25, 2012

### Staff: Mentor

One of the 100Ω is shorted out. Total R is 40||25||25||100, in series with the 10Ω

6. Jan 25, 2012

### Curious3141

You're absolutely right. Missed that, thanks for spotting it.

Still doesn't give him a "nice" answer though - but at least it'll be the right one.

7. Jan 26, 2012

### TheRealDoodle

So if im understanding this right, because our teacher hasnt taught us the (1/r1+...) method yet, I will do (1/40 + 1/25 +1/25 + 1/100) which will give me 0.115, then added to the 10ohm resistor will be 10.115, and since I=V/R then 10/10.115= 988.6mA which I know isnt right and I must be doing something wrong because I saw the right answer on my friends homework when he got it back today and he wrote 524mA

8. Jan 26, 2012

### Staff: Mentor

Yes. That gives you 1/R, so invert it to find R. The formula is 1/R = 1/r1 + 1/r2 + ....

Then add 10, and you have the total circuit resistance.

9. Jan 26, 2012

### Curious3141

The equation for determining the effective resistance in a parallel arrangement of n resistors is:

$$\frac{1}{R_{eff}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + ... + \frac{1}{R_{n}}$$

So you have to take the reciprocal (inverse) of that sum to get the effective resistance, which is 1/0.115 = 8.696Ω. This has to be added to the 10Ω in series to get 18.696Ω, which is the total resistance in the circuit.

The current is therefore 10V/18.696Ω = 0.535A = 535mA, which should be the right answer.

The 524mA is using the "200" figure instead of the "100" figure as I originally did. This is a mistake, because one of the 100Ω is shorted out. It's easy to miss it because the short is "after" it (distal to it).