Kirchoff's rules & resistors in parallel

[Flucky]

Homework Statement

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)

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?

Thanks in advance

 

[bsbs]

Mistake in your second law,
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.

 

[Flucky]

Mistake in your second law,
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.

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

 

[bsbs]

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

Loop 1,

-R₁(I₂)+E=0

Loop 2,

-R₂(I₁-I₂)+R₁(I₂)=0

 

[Flucky]

i help you with one...try to write the other loop

Loop 1,

-R₁(I₂)+E=0

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 ?

 

[bsbs]

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 ?

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

ok? solved?

 

[Flucky]

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

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

 

[superdave]

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

R = (1/R₁ + 1/R₂)^-1

What are you getting for R?

 

[bsbs]

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

sorry, didnt notice that your eq is also correct.

 

[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?