# 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.
Homework 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?

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#### berkeman

Mentor
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 Helper
Gold Member
2018 Award
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?

Last edited:

#### haruspex

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Gold Member
2018 Award
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

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Gold Member
2018 Award
$R_2I_2$ 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

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Gold Member
2018 Award
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

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2018 Award
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

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2018 Award
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

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Gold Member
2018 Award
+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

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!