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Finding the current and voltage for a resistor

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Homework Statement
For the circuit in Figure 1, find the current and voltage at resistor R1 considering that the values of the voltage sources
they are V1 [V] and V2 [V], and the resistance values are R1 [Ω], R2 [Ω] and R3 [Ω] respectively.
Homework Equations
Kirchhoff's Laws:
##\sum_i I_i=0##
##\sum_i V_i=0##
1573408396353.png

I tried to solve it by loop currents. So on the left mesh the loop current ##I_1## goes clockwise and on the right mesh the loop current ##I_2## goes counterclockwise.

1573409337462.png

I ended up with the following equations:
1) ##V_1-R_1(I_1+I_2)-R_2I_1=0##;
2) ##V_2-R_3I_2-R_1(I_1+I_2)=0##.
To find the current on ##R_1##, I tried to find ##I_1## and ##I_2##. I first solved 1) for ##I_1##: ##I_1=\frac{V_1-R_1I_2}{R_1+R_2}##. Then I solved 2) for ##I_2## replacing the expression that I found for ##I_1## :

##V_2-R_3I_2-R_1(I_1+I_2)=V_2-I_2(R_3-R_1)-R_1I_1##
##=V_2-I_2(R_3-R_1)-R_1\left(\frac{V_1-R_1I_2}{R_1+R_2}\right)=V_2-I_2(R_3-R_1)-R_1\left(\frac{V_1-R_1I_2}{R_1+R_2}\right)##
##=V_2(R_1+R_2)-I_2(R_3-R_1)(R_1+R_2)-R_1(V_1-R_1I_2)##
##=V_2(R_1+R_2)-I_2[(R_3-R_1)(R_1+R_2)-R_1^2]-R_1V_1##
##I_2=\frac{V_2(R_1+R_2)-R_1V_1}{(R_1+R_2)(R_3-R_1)-R_1^2}##​
Finally, I replaced ##I_2## on ##I_1##: ##I_1=\frac{V_1-R_1\left(\frac{V_2(R_1+R_2)-R_1V_1}{(R_1+R_2)(R_3-R_1)-R_1^2}\right)}{R_1+R_2}##. Am I correct?

I'm not sure how to find the voltage on ##R_1##, but I think that I need to find ##R_1(I_1+I_2)##, right?
 

berkeman

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TBH, I never liked working problems with KVL. Can you try solving it with KCL instead? My KCL equations for this problem are a lot simpler than what you posted...
 

haruspex

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Homework Statement: For the circuit in Figure 1, find the current and voltage at resistor R1 considering that the values of the voltage sources
they are V1 [V] and V2 [V], and the resistance values are R1 [Ω], R2 [Ω] and R3 [Ω] respectively.
Homework Equations: Kirchhoff's Laws:
##\sum_i I_i=0##
##\sum_i V_i=0##

View attachment 252658
I tried to solve it by loop currents. So on the left mesh the loop current ##I_1## goes clockwise and on the right mesh the loop current ##I_2## goes counterclockwise.

View attachment 252659
I ended up with the following equations:
1) ##V_1-R_1(I_1+I_2)-R_2I_1=0##;
2) ##V_2-R_3I_2-R_1(I_1+I_2)=0##.
To find the current on ##R_1##, I tried to find ##I_1## and ##I_2##. I first solved 1) for ##I_1##: ##I_1=\frac{V_1-R_1I_2}{R_1+R_2}##. Then I solved 2) for ##I_2## replacing the expression that I found for ##I_1## :

##V_2-R_3I_2-R_1(I_1+I_2)=V_2-I_2(R_3-R_1)-R_1I_1##
##=V_2-I_2(R_3-R_1)-R_1\left(\frac{V_1-R_1I_2}{R_1+R_2}\right)=V_2-I_2(R_3-R_1)-R_1\left(\frac{V_1-R_1I_2}{R_1+R_2}\right)##
##=V_2(R_1+R_2)-I_2(R_3-R_1)(R_1+R_2)-R_1(V_1-R_1I_2)##
##=V_2(R_1+R_2)-I_2[(R_3-R_1)(R_1+R_2)-R_1^2]-R_1V_1##
##I_2=\frac{V_2(R_1+R_2)-R_1V_1}{(R_1+R_2)(R_3-R_1)-R_1^2}##​
Finally, I replaced ##I_2## on ##I_1##: ##I_1=\frac{V_1-R_1\left(\frac{V_2(R_1+R_2)-R_1V_1}{(R_1+R_2)(R_3-R_1)-R_1^2}\right)}{R_1+R_2}##. Am I correct?

I'm not sure how to find the voltage on ##R_1##, but I think that I need to find ##R_1(I_1+I_2)##, right?
You have a sign error in ##V_2-R_3I_2-R_1(I_1+I_2)=V_2-I_2(R_3-R_1)-R_1I_1##.
An easier way to proceed from (1) and (2) is to find a linear combination of them such that all references to currents are in the form ##I_1+I_2##.
 
393
13
TBH, I never liked working problems with KVL. Can you try solving it with KCL instead? My KCL equations for this problem are a lot simpler than what you posted...
That would just change the ##I_1+I_2##, right? It would give an equation of ##I_1+I_2=I_3## I think.
 
393
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You have a sign error in ##V_2-R_3I_2-R_1(I_1+I_2)=V_2-I_2(R_3-R_1)-R_1I_1##.
An easier way to proceed from (1) and (2) is to find a linear combination of them such that all references to currents are in the form ##I_1+I_2##.
You're right, it would be ##V_2-R_3I_2-R_1(I_1+I_2)=V_2-I_2(R_3+R_1)-R_1I_1## , right?

How do I find a linear combination?
 
Last edited:

haruspex

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You're right, it would be ##V_2-R_3I_2-R_1(I_1+I_2)=V_2-I_2(R_3+R_1)-R_1I_1## , right?

How do I find a linear combination?
It just means multiplying each equation by some variable (different ones) and adding them.
In the equations, some terms are already in the form I1+I2, but in one you have an R2I1 and the other an R3I2. What two values do you need to multiply by so that both currents have the same coefficient?
 
393
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It just means multiplying each equation by some variable (different ones) and adding them.
In the equations, some terms are already in the form I1+I2, but in one you have an R2I1 and the other an R3I2. What two values do you need to multiply by so that both currents have the same coefficient?
##R_2I_2## would need to be multiplied by ##\left(\frac{1}{R_2}+\frac{I_2}{R_2I_1}\right)## and ##R_3I_2## by ##\left(\frac{1}{R_3}+\frac{I_1}{R_3I_2}\right)##, am I correct?
 

haruspex

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##R_2I_2## would need to be multiplied by ##\left(\frac{1}{R_2}+\frac{I_2}{R_2I_1}\right)## and ##R_3I_2## by ##\left(\frac{1}{R_3}+\frac{I_1}{R_3I_2}\right)##, am I correct?
You mean ##R_2I_1##, not ##R_2I_2##. But no, it's much simpler than that. A single variable in each case.
 
393
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You mean ##R_2I_1##, not ##R_2I_2##. But no, it's much simpler than that. A single variable in each case.
Yes, sorry for the typo. How would it be? Maybe ##R_3## and ##R_2##? then both would have ##R_2R_3##.
 

haruspex

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Yes, sorry for the typo. How would it be? Maybe ##R_3## and ##R_2##? then both would have ##R_2R_3##.
Right. That should give you an equation from which you can find ##I_1+I_2## immediately.
 
393
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Right. That should give you an equation from which you can find ##I_1+I_2## immediately.
Ok, then I would have:
##V_1R_3-R_1R_3(I_1+I_2)-R_2R_3I_1=0##
##V_2R_2-R_1R_2(I_1+I_2)-R_2R_3I_2=0##
Then I can subtract, right?
 

haruspex

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Ok, then I would have:
##V_1R_3-R_1R_3(I_1+I_2)-R_2R_3I_1=0##
##V_2R_2-R_1R_2(I_1+I_2)-R_2R_3I_2=0##
Then I can subtract, right?
No, if you subtract you will have a term ##R_2R_3(I_1-I_2)##.
 
393
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No, if you subtract you will have a term ##R_2R_3(I_1-I_2)##.
Then I'll add them, obtaining ##V_1R_3+V_2R_2-R_1R_3(I_1+I_2)-R_1R_2(I_1+I_2)-R_2R_3I_1-R_2R_3I_2=0## thus ##V_1R_3+V_2R_2-(I_1+I_2)[R_1R_3+R_1R_2]-R_2R_3(I1+I_2)=0##
##\Rightarrow V_1R_3+V_2R_2-(I_1+I_2)[R_1R_3+R_1R_2+R_2R_3]=0##. Am I correct?
 

haruspex

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Then I'll add them, obtaining ##V_1R_3+V_2R_2-R_1R_3(I_1+I_2)-R_1R_2(I_1+I_2)-R_2R_3I_1-R_2R_3I_2=0## thus ##V_1R_3+V_2R_2-(I_1+I_2)[R_1R_3+R_1R_2]-R_2R_3(I1+I_2)=0##
##\Rightarrow V_1R_3+V_2R_2-(I_1+I_2)[R_1R_3+R_1R_2+R_2R_3]=0##. Am I correct?
That looks better.
 
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+I
That looks better.
Then I only have to solve for ##I_1+I_2## and to find the voltage on ##R_1## I just multiply ##I_1+I_2## times ##R_1##, right?
 

haruspex

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Thank you. I guess that if I continued simplifying my original expressions for ##I_1## and ##I_2## nd adding them I would arrive to the same answer, but this helped me a lot. Thanks!
 

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