# Current in 10 ohm Resistor

1. Feb 13, 2012

### KendrickLamar

1. The problem statement, all variables and given/known data
Find the current in the 10.0 ohm resistor in the drawing (V1 = 24.0 V and R1 = 15.0 Ohm)

Diagram:

2. Relevant equations
Kirchoffs loop rule

3. The attempt at a solution

well do i have to solve for both parts of the circuit if i just want to find the current at the 10.0 ohm resistor or can i just do one side?

could i do like 10 - R1*I1 - 10*(I1-I2) + 10 = 0?

well actually then i need to find I2 right? or how can i solve it, like am i just looking for I1-I2? or what im just confused at this point

2. Feb 13, 2012

### SammyS

Staff Emeritus
You pretty much answered your own question. Using only one loop leaves you with two unknowns in one equation. You also need to do the other loop.

3. Feb 13, 2012

### Mindscrape

Yeah, you need another loop. Since when do rappers take physics? :p

4. Feb 13, 2012

### KendrickLamar

how do you set up the 2nd loop?

V1 + 10 - 10*(I2-I1)? - 5.0*I2 ??

or is the bolded part +10*(I2-I1) or is it I2 + I1 ? or what? and does the rest of that equation look right?

5. Feb 13, 2012

### Mindscrape

Yeah, you've got it right. Solve the simultaneous eqns.

6. Feb 13, 2012

### KendrickLamar

well for the first loop the +10 from the first voltage would cancel out with the one to the right of it correct?

and am i looking for I2-I1 or I1-I2 when looking for the current in the 10 ohm resistor?

7. Feb 13, 2012

### KendrickLamar

Alright can someone double check this?

so i did for the first loop
10 - 15I1 - 10I1 + 10I2 - 10 = 0
25I1 = 10I2

then 2nd loop did
24 + 10 - 10(I2-I1) - 5I2 = 0
simplified it to => 34 = 15I2-10I1 and then to => 27.5*I1 = 34

plugged I1 in for I2 from the first equation and ended up with
I1 = 1.236
I2 = 3.09

Plugged it in now to 10 ohms * (I2 - I1) = 10*(3.09-1.236) = 18.55 amps

is that the right answer? and i only did I2-I1 because I2 was bigger but im not sure it should matter? i dont know lol im confused on this part

edit: i def. did something wrong lol it should NOT be that high

Last edited: Feb 13, 2012
8. Feb 13, 2012

### SammyS

Staff Emeritus
If you're looking for the current through the 10Ω resistor, why did you multiply current by resistance? --- That gives the voltage drop across the 10Ω resistor.

9. Feb 13, 2012

### KendrickLamar

wait oh yeah, but the Current through that point I2-I1 or I1-I2?

10. Feb 13, 2012

### SammyS

Staff Emeritus
It's possible to do it either way.

Which way makes sense to you and why?