- #1
Knissp
- 72
- 0
[SOLVED] Kirchoff's Laws
A link to the problem statement is provided here: http://img231.imageshack.us/img231/5305/kirchoffry1.png
(with one modification: the question should read find the current in the .75 ohm resistor)
Ohm's Law: R=delta(V)/I
loop rule: sum(changes in potentials) = 0 (for closed loop)
junction rule: because of conservation of charge, the currents follow such that:
I_1 + I_2 = I_3
I am supposed to solve a set of simultaneous linear equations. The first one is simply the node equation
I_1+I_2=I_3.
Then by looking at the top loop, I have:
2.25 - (I_2)*(3) - (I_3) * (4.2) = 0
The next part is what I am somewhat confused about.
I can either make one of two loops for the final equations: the bottom one with the middle wire included, or the top and bottom one as a whole.
For just the bottom loop, I get:
4.75 - (I_1)*(.75) + 1 - (I_1)*(1.2) + (I_2)*(3) +2.25 = 0
I am not sure if this is right because I don't know if it is appropriate to add the voltages 1 and 4.75 when they have the resistor between them. Perhaps I am supposed to use their potential difference? But then in that case, the .75 ohms would be internal resistance and I don't have any emf. And, I think it is probably incorrect to assume that the current going through the 1.2 ohm resistor is the same as I_1.
For the whole loop (the biggest one which does not include the middle wire), I get:
4.75 - (I_1)*(.75) + 1 - (I_1)*(1.2) - (I_3)*(4.2) = 0
When I solve this simultaneously with the first two equations, I get a different answer from when I solve the first three simultaneously.
Homework Statement
A link to the problem statement is provided here: http://img231.imageshack.us/img231/5305/kirchoffry1.png
(with one modification: the question should read find the current in the .75 ohm resistor)
Homework Equations
Ohm's Law: R=delta(V)/I
loop rule: sum(changes in potentials) = 0 (for closed loop)
junction rule: because of conservation of charge, the currents follow such that:
I_1 + I_2 = I_3
The Attempt at a Solution
I am supposed to solve a set of simultaneous linear equations. The first one is simply the node equation
I_1+I_2=I_3.
Then by looking at the top loop, I have:
2.25 - (I_2)*(3) - (I_3) * (4.2) = 0
The next part is what I am somewhat confused about.
I can either make one of two loops for the final equations: the bottom one with the middle wire included, or the top and bottom one as a whole.
For just the bottom loop, I get:
4.75 - (I_1)*(.75) + 1 - (I_1)*(1.2) + (I_2)*(3) +2.25 = 0
I am not sure if this is right because I don't know if it is appropriate to add the voltages 1 and 4.75 when they have the resistor between them. Perhaps I am supposed to use their potential difference? But then in that case, the .75 ohms would be internal resistance and I don't have any emf. And, I think it is probably incorrect to assume that the current going through the 1.2 ohm resistor is the same as I_1.
For the whole loop (the biggest one which does not include the middle wire), I get:
4.75 - (I_1)*(.75) + 1 - (I_1)*(1.2) - (I_3)*(4.2) = 0
When I solve this simultaneously with the first two equations, I get a different answer from when I solve the first three simultaneously.
Last edited by a moderator:
