Resistance in parallel circuits! :D

Hey!! this question will seem really stupid haha but i need some help...

I know that

if you have e.g, two 4 Ohm resistors in parallel, its the same as having one 2 ohm resistor in series.

Because e.g. if current is 3 in each circuit...

1.5 * 4 = 6 volts in the parallel circuit across each component.
3 * 2 = 6 volts in series circuit..

i know that...but i can only explain using values...

theres a 2 mark question in an exam paper saying...

'Explain why two identical resistors connected in parallel have a combined resistance of half the value of one resistor on its own'

I couldnt write all of that in the line they gave....

how can you answer that question with a more general rule? I understand it...with figures...but i cant write it out without an example...

please help thanks all

Reply 1

Adjective

\displaystyle \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2}

. Find what happens when R1 = R2.

If you're not happy with where this formula comes from, say so and I'll show you.

Reply 2

lilpenguin

Thanks, il try it out..

1/R1 + 1/R1 = 2/R1

2/R1 = 1 / Rt?

2Rt = 1R1?

So 2 parallel resistances are the same as 1 R1 resistor?

Ahh i think thats wrong, could you explain? :biggrin:

Reply 3

Kengurnu

basically if you think of a big stadium full of people, all trying to get out:

Open one small door (one 4ohm) and out they come slowly. open 2 small doors (2x 4ohm in parallel) and they come out twice as fast. open one big door, twice the width of a small door (one 2ohm), and they come out at the same speed as if you'd opened two small doors.

That's the best way I've found of explaining the concept, but other than that I know what you mean, it's hard to prove algebraically!

Reply 4

Adjective

You're right, but you haven't divided by 2 at the end!

Here's what you did: Let's say each resistor has resistance x

\Omega

. So

\displaystyle \frac{1}{R_T} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x}

Therefore

\displaystyle R_T = \frac{x}{2} \Omega

Reply 5

lilpenguin

Kengurnu, yeah! :P That makes sense, its just hard to explain etc as you stated! :biggrin:

Adje, ohh i think im beginning to understnad, 2/R1 = 1/Rt

Rt = R1/2
which is half..?
:P instead of multiplying out just flip the other bracket? :biggrin:

thanks for help btw every one :biggrin:

Adje, do you have Jan 07 Physics in action question available to you by any chance?

I apply all the things youve just taught me, and tried the V=IR Formula on a simple 2 mark question about Voltage in a circuit, and it wont work! Do you think you could help??

For this question here, i unerstand what you said, it really helps :biggrin:

i just need practise applying it :biggrin:

Reply 6

lilpenguin

Jan 07, Physics in action, Question 9, part B ii part 2.

Its got a 12 V supply, with two 25kO resistors kinda in series...
but across one of them, is a Voltmeter with 25kO resistance also...

and it says find the voltage across AB,

(which is across the resistor which has the voltmeter across it)

The answer is 4 Volts, ive tried so many ways, using the 1/R thing as the Voltmeter is in parallel? but i just cant do it! :frown:

its one of those questions its only 2 marks but im going insane :biggrin:

Reply 7

Adjective

With regards to your

R_T

question, in effect, yes you just flip the fractions round on both sides.

Spoiler

Gimme a sec and I'll do that question.

Reply 8

lilpenguin

Thank you! :P That makes alot of sense.

And thank you again! :biggrin:

Im not sure how to do it because its kinda half series half parallel circuit :redface:

im so bad at circuits haha :biggrin:

Reply 9

Adjective

Right, so the circuit is like this? If so, assuming no internal resistance in the cell, the voltage should be 6V, since both resistors are identical...

dfs.jpg12.2KB

Reply 10

lilpenguin

The circuit is like that..
except there is 25kO across the other resistor also and also 25kO resistance in the voltmeter

yeah the diagram is correct im not sure if the other resistances need to be included in a calculation?

yeah i thought 6V too but mark scheme says 4V..
:frown:

Reply 11

Adjective

Ah, okay, so there are three 25kO resistances - two in the resistors and one in the voltmeter.

In that case:

The Long Way:
First, we find out the combined resistance of the voltmeter and the resistor it's attached across:

Since the individual resistances are the same, we know (from the stuff we've just done) that the combined resistance is half of one individual resistance... i.e. 12.5kO.

The other resistor still has a resistance of 25kO.

V = IR

We have to find the current. The resistance is (25 + 12.5)kO, and the voltage is 12V. So the current is

\displaystyle \frac{12}{25000 + 12500} = 3.2 \times 10^{-4} A

.

Since the resistance of the voltmeter and the resistor are the same, the current splits equally between them - so the current is

1.6 \times 10^{-4} A

.

V = IR
V =

1.6 \times 10^{-4} \, \, \times 25000 = 4V

The Shorter Way:

First, we find out the combined resistance of the voltmeter and the resistor it's attached across:

Since the individual resistances are the same, we know (from the stuff we've just done) that the combined resistance is half of one individual resistance... i.e. 12.5kO.

Now we treat the combined resistance as a single resistor.
The ratio of the voltage across the 12.5kO resistance to the voltage across the 25kO resistance is the same as the ratio of the 12.5kO resistance to the 25kO resistance - i.e.

\displaystyle \frac {V_{12.5k}}{V_{25k}} = \frac {12500}{25000} = 0.5

. So you know that the voltage across the resistor attached to the voltmeter is half the voltage of the 25k resistor.

You also know that the two voltages have to add up to 12. What are the only two numbers that fulfil both those requirements? 4 and 8. So the voltages across the resistor, and the voltage across the voltmeter are both 4V.

Reply 12

lilpenguin

1 word.. Sensational!

That is so great, that 'long' method is great it just goes through all the steps :biggrin:

so many little things :P thank you SO much for your help, you have no idea how appreciated it is! :biggrin:

Reply 13

Adjective

Haha, my pleasure - helps me revise for PHY2 (on Thursday)! You might find the 'shorter' way (edited post) a bit more straightforward... depends how you think I suppose...