# Homework Help: Kirchhoff's Rules

1. Apr 18, 2014

### BOAS

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...

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

Kirchhoff's junction and loop rule.

3. 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 cant find the mistake.

BOAS

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2. Apr 18, 2014

### Staff: Mentor

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.

3. Apr 18, 2014

### Tanya Sharma

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.

4. Apr 18, 2014

### BOAS

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.

5. Apr 18, 2014

### Tanya Sharma

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.

6. Apr 18, 2014

### BOAS

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.