Circuit Question

Five identical resistors are connected to a cell in the diagram (attached). The potential differences across R1, R2 and R3 are V1, V2 and V3 respectively. What is the order of increasing potential differences, smallest first?

- If I sub in values for the resistors and for the voltage of the battery, I still don't know how to arrive at the correct answer. I know the voltage is the same in a parallel circuit, so R1 received half of the voltage from the battery, and R1 and R2 have the same resistance. My only problem is with the parallel combination with R3; I mean I know the total resistance of the combination is less than the resistance of each resistor, but just don't know how to answer the question.

Any guidance please? Hope you can see my problem

As the resistors are equal call them all R ohms. The two in parallel add to equal one resistor of value ½R (from 1/R = 1/R1 + 1/R2)

The circuit is now



The resistance between A and B is less than that between C and D. Can you see why?
So the current I2 is greater than the current I1
The pd across any resistor is I times R so
-the pd across the top 2 resistors is the same. What about the bottom 2? How does the first one (pd across R) on the bottom compare with the top R?
The pd across R3 in the question is the same as that across R/2 in my diagram.
The pd across (R/2) will be half that across the other one between C and D.

Thanks for the reply Surely the resistance between A and B is greater than that between C and D, hence I2 is greater than the current I1, no?

The first one (pd across R) on the bottom has a greater pd than the pd on the top R, as they both have the same resistance but the one below has greater current hence greater pd.
Would the pd across R3 in the question not be half of the pd across R/2 in your diagram since there are 2 branches?
I understand that part, thanks

Oops that was a typo. Yes it should have (and now does) read that the reistance between A and B is greater than...
The conclusion was ok though. I2 > I1
It means that the V2 must be greater than V1 because it's the same resistance in the two cases but a larger current through R2.
And V=IR
The pd V3 across the equivalent resistor R/2 will be the pd across both resistors in parallel in the original diagram.
This pd must be half that across R2 because the equivalent resistance there is half that of R2

No worries
And right I see, so the pd across the resistor R/2 = the pd across the resistor R3 on my diagram as pd is always equal in parallel, correct?

Yes V3 is half V2 and V3 in my diagram is the pd across R3 and also across the other resistor in parallel with it.

Hi,

Sorry to come back to an old post but you see the resistance in the diagram you have R/2 - i thought you work out resistance in parallel circuits to be 1/R so 1/R+1/R = 0.5R

Correct. But you missed out the step to get to that answer:

Adding the fractions:

1/R + 1/R = (1 + 1)/R = 2/R

And the inverse is R/2

As the resistors are equal call them all R ohms. The two in parallel add to equal one resistor of value ½R (from 1/R = 1/R1 + 1/R2)

The circuit is now



The resistance between A and B is greater than that between C and D. Can you see why?
So the current I2 is greater than the current I1
The pd across any resistor is I times R so
-the pd across the top 2 resistors is the same. What about the bottom 2? How does the first one (pd across R) on the bottom compare with the top R?
The pd across R3 in the question is the same as that across R/2 in my diagram.
The pd across (R/2) will be half that across the other one between C and D.

Hi but why is R3 smaller than R1 though?