Solving Kirchhoff's Rules: Find Currents in a Circuit w/2 Batteries

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The discussion revolves around solving a circuit problem using Kirchhoff's rules, specifically involving two batteries and three resistors. The user initially struggles to solve the equations derived from the circuit, which include relationships for currents and voltage drops. Other participants confirm that the equations are correctly set up and suggest using substitution to simplify the problem. The user realizes the need to practice simultaneous equations and successfully finds the current across resistor R2, determining it flows in the opposite direction than initially assumed. The conversation highlights the importance of careful equation manipulation in circuit analysis.
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Hi,

i'm stuck on this problem and have been trying to identify where I might have made an error but have come up empty. I have assigned my loops and junctions and have three equation for three unknowns, but can't seem to solve them...

Homework Statement



Consider the circuit in the following diagram (attached):

where the two ideal dc batteries deliver a terminal voltage of V1 = 2V and V2 = 10V respectively. The magnitude of the resistances is R1 = 2Ω; R2Ω = 2; and R3 = 3Ω. Calculate the magnitude and direction of the current across the resistor R2.

Homework Equations



Kirchhoff's junction and loop rule.

The Attempt at a Solution



I1 = I2 + I3 (JR)

v1 - I1R1 - I2R2 = 0 (loop A)

-I3R3 + V2 + I2R2 = 0 (loop B)

I can't see how to solve for any of these, which is making me think I've worked out the expressions wrong, but I can't find the mistake.

Thanks for any help you can give,

BOAS
 

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Your equations look fine. Must be something to do with your mechanics of solving the equations. Guess you'll have to provide some details there.
 
BOAS said:
I1 = I2 + I3 (JR)

v1 - I1R1 - I2R2 = 0 (loop A)

-I3R3 + V2 + I2R2 = 0 (loop B)

Your work looks fine.

You have three equations and three unknown so there should be no problem .From first equation find the value of i3 and substitute into 2nd and 3rd equations .You will be left with two equations and two unknowns.
 
Tanya Sharma said:
Your work looks fine.

You have three equations and three unknown so there should be no problem .From first equation find the value of i3 and substitute into 2nd and 3rd equations .You will be left with two equations and two unknowns.

I3 = I1-I2

Subbing into the 3rd equation gives;

V2 + I2R2 - (I1-I2)R3 = 0

I don't see what you meant, when you said to sub I3 into the second equation.
 
BOAS said:
v1 - I1R1 - I2R2 = 0 (loop A)

BOAS said:
V2 + I2R2 - (I1-I2)R3 = 0

Now you have two equations and two unknowns i1 and i2.The values of V1,V2,R1,R2.R3 are given in the problem statement. Just solve the two equations.
 
Tanya Sharma said:
Now you have two equations and two unknowns i1 and i2.The values of V1,V2,R1,R2.R3 are given in the problem statement. Just solve the two equations.

I've been making this far hard than it needs to be - I definitely need to go back and practice simultaneous equations again. I had completely forgotten to multiply the equations by a number to make elimination possible.

Multiplying V1 - I1R1 - I2R2 = 0 by 3/2 makes the elimination work.

I get an answer for I2 = -7/8 A which means the current goes the other way round.

Thank you for your help.
 
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