# Kirchoff's rules & resistors in parallel

1. Nov 16, 2012

### Flucky

1. The problem statement, all variables and given/known data
Use Kirchhoff’s junction rule and loop rule (twice) to find the effective resistance of the two
resistors connected in parallel as shown, and demonstrate that this agrees with the expression 1/R = 1/R1 + 1/R2

(poorly drawn picture of the circuit)

3. The attempt at a solution

Using Kirchoff's 1st law:
I1 = I2 + I3

Using Kirchoff's 2nd law:
E - I2R1 = 0 ........ [1]
and
I2R1 - I3R2 = 0 ........ [2]

Now I think I can say that E = I1R
Subtituting that into equation [1] gives:
I1R - I2R1 = 0

Now here's where I'm having difficulty, I keep trying out different ways of substituting the equations into one another but end up with stupid answers like R = R2...

Could anyone point me in the right direction?

2. Nov 16, 2012

### bsbs

This is not a complicated circuit, just use two variables for current if u want.

Replace i3 as (i1-i2)

See the circuit as two individual boxes, running your finger from the top left hand corner and try to apply the rule again.

Last edited: Nov 16, 2012
3. Nov 16, 2012

### Flucky

Sorry I must be missing something, I can't see what's wrong with the 2nd law bit

4. Nov 16, 2012

### bsbs

Loop 1,

-R1(I2)+E=0

Loop 2,

-R2(I1-I2)+R1(I2)=0

Last edited: Nov 16, 2012
5. Nov 16, 2012

### Flucky

That's what I got for my loop 1, so it must be my loop 2 that's incorrect.

If it's not I2R1 - I3R2 = 0 is it E - I2R3 = 0 ?

6. Nov 16, 2012

### bsbs

see my loop 2 above, also take note that your E=IR

ok? solved?

Last edited: Nov 16, 2012
7. Nov 16, 2012

### Flucky

Thanks for replying but we're back to the beginning now as that's what I had already got, it's the next bit I'm having trouble with - getting the equation 1/R = 1/R1 + 1/R2

8. Nov 16, 2012

### superdave

R = (1/R1 + 1/R2)-1

What are you getting for R?

9. Nov 16, 2012

### bsbs

sorry, didnt notice that your eq is also correct.

Last edited: Nov 16, 2012
10. Nov 17, 2012

### bsbs

have you tried substitute ohm's law into your KCL?

KCL states :current at a junction is zero
ohm states: I=E/R

something like I1=E/(1/R1 + 1/R2)-1

etc etc....does it helps?