Voltage Drop across multiple resistors

[talaroue]

Homework Statement

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

Homework Equations

V=IR

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 don't understand to find the voltages now...
Do I simply just do (I1*R1)-Vs=V1?

 

[Doc Al]

I don't understand to find the voltages now...
Do I simply just do (I1*R1)-Vs=V1?

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?

 

[talaroue]

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

 

[Doc Al]

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

Then just find the voltage drops across each resistor.

 

[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.

 

[Doc Al]

which is just R*I=V right?

Yes.

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?

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.

 

[talaroue]

Solving for I's...

i_t=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 i_t=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 don't have to subtract anything?

 

[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?

 

[Doc Al]

Solving for I's...

i_t=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 i_t=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

Looks good.

thats all?
I don't have to subtract anything?

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.)

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?

Right.

 

[talaroue]

I am starting to understand this! He doesn't really state but on other problems he has a voltage for the resistor listed. just like on one of my other problems.