How Do You Calculate Current in a 10 Ohm Resistor Using Kirchhoff's Loop Rule?

[KendrickLamar]

Homework Statement

Find the current in the 10.0 ohm resistor in the drawing (V1 = 24.0 V and R1 = 15.0 Ohm)

Diagram:

Homework Equations

Kirchoffs loop rule

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 I am just confused at this point

 

[SammyS]

Homework Statement

Find the current in the 10.0 ohm resistor in the drawing (V1 = 24.0 V and R1 = 15.0 Ohm)

Diagram:

Homework Equations

Kirchoffs loop rule

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 I am just confused at this point

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.

 

[Mindscrape]

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

 

[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?

 

[Mindscrape]

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

 

[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?

 

[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 I am not sure it should matter? i don't know lol I am confused on this part

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

 

[SammyS]

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 I am not sure it should matter? i don't know lol I am confused on this part

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

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.

 

[KendrickLamar]

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.

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

 

[SammyS]

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

It's possible to do it either way.

Which way makes sense to you and why?