# Voltage Drop across multiple resistors

1. Aug 31, 2009

### talaroue

1. The problem statement, all variables and given/known data
Solve for all unkown i's and V's
EDIT: there is a seventh resistor that is right under the negative side of the source.

Vs=12V

2. Relevant equations
V=IR

3. The attempt at a solution

I got the I's by finding the voltage across the entire 2 loops to equal 0, then just simply had 3 equations, and 3 uknowns and went from there and got them to be

It=48mA I1=24mA I2=24mA

I dont understand to find the voltages now.....
Do I simply just do (I1*R1)-Vs=V1?

2. Aug 31, 2009

### Staff: Mentor

It's not clear to me what you are asked to find. The voltage drop across each resistor? Or the voltage at certain points with respect to some reference?

3. Aug 31, 2009

### talaroue

All it asks for is Solve for all unkown i's and V's

4. Aug 31, 2009

### Staff: Mentor

Then just find the voltage drops across each resistor.

5. Aug 31, 2009

### talaroue

which is just R*I=V right? It just doesn't make sense that some times I have to have a different I for each resistor and then sometimes like this one I only need 2. for the whole circuit. Do you understand what I mean?

I will post how I found I my next post.

6. Aug 31, 2009

### Staff: Mentor

Yes.
Realize that current must be the same throughout a single branch of the circuit. There are only three independent branches, thus only three currents needed to describe this circuit. Also, since the resistors are all the same, there'll only be three different voltage drops.

7. Aug 31, 2009

### talaroue

Solving for I's................

it=i1+i2

with i1 going to the loop to the left, and i2 going to the loop to the right.

Where I came up with the equations......
12-R1i1-R6i1-R5i1-R7i=0

12-R2i2-R3i2-R4i2-R7i=0

then plugging in it=i1+i2 for in in both and then just combinding like terms I got these two equations

12-400i1-100i2=0
12-100i1-400i2=0
multiplying the top equation by (-4) I2 cancels out leaving just I1 so i just solved and got

I1=36/1500=24mA then plugged that back into the equations about to get I2 I got 24mA for I2 as well.

I=I1+I2=24mA+24mA=48mA

so then for the voltage across each resistor is simply the corresponding I to the resistor for example for R1......

V=IR
V=?
I=I1=24mA
R=100 Ohms

thats all?
I dont have to subtract anything?

8. Aug 31, 2009

### talaroue

So basically the only time a new current is created, i guess you could say is if it splits from one to 2 or more?

Why don't you have to subtract anything?

9. Aug 31, 2009

### Staff: Mentor

Looks good.
It depends on what is meant by "find all the unknown Vs". Does your instructor usually take a particular point as being 0 V? (For example, on one side of the voltage supply. But that's arbitrary.)

Right.

10. Aug 31, 2009

### talaroue

I am starting to understand this! He doesnt really state but on other problems he has a voltage for the resistor listed. just like on one of my other problems.