Finding the current and voltage for a resistor

[Davidllerenav]

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.

Relevant Equations

Kirchhoff's Laws:
$\sum_i I_i=0$
$\sum_i V_i=0$

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.

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]

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]

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 $I_1+I_2$.

 

[Davidllerenav]

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.

 

[Davidllerenav]

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?

 

[haruspex]

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?

 

[Davidllerenav]

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]

$R_2I_2$ 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.

 

[Davidllerenav]

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]

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.

 

[Davidllerenav]

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]

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 $R_2R_3(I_1-I_2)$.

 

[Davidllerenav]

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]

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.

 

[Davidllerenav]

+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]

+I
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?

Yes.

 

[Davidllerenav]

Yes.

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!