# Weird Circuit Multiple Choice Question??Watch

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#1
Ahhh ive spent about an hour trying different combinations of resistances to try and get the correct answer for this question, but I just can't seem to get the correct answer.
The correct answer is 2R (B)
0
3 years ago
#2
I think it's A Level's Question as I have never solved this type of in my O Levels. Sorry, can not help you.
0
3 years ago
#3
(Original post by HasanRaza1)
Ahhh ive spent about an hour trying different combinations of resistances to try and get the correct answer for this question, but I just can't seem to get the correct answer.
The correct answer is 2R (B)

I tried to redraw the circuit as shown below.

I believe you can calculate the effective resistance in the series and parallel set up.
0
3 years ago
#4
(Original post by HasanRaza1)
Ahhh ive spent about an hour trying different combinations of resistances to try and get the correct answer for this question, but I just can't seem to get the correct answer.
The correct answer is 2R (B)
You've got to break the circuit down into bits you can handle.

What are really easy to handle are circuits containing nothing but resistors in series and nothing but resistors in parallel.

first the two resistors on the right hand side (3R and 6R)... those are a pair of resistors in parallel which can be replaced with a single resistor using the rule for parallel resistors. Reffective = (1/R1 + 1/R2)-1

now the circuit looks like this...

and you've got 3 resistors in series that you can replace with one...

Attachment 531351531355

which you know how to solve as 2 resistors in parallel
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#5
(Original post by Eimmanuel)
I tried to redraw the circuit as shown below.

I believe you can calculate the effective resistance in the series and parallel set up.
Thanks a lot for doing that )
However, when i look at this setup, I see that in the first branch with the 4 ohm resistor the resistance is 4R
In the second branch, the resistance is R plus R, plus 0.5R (due to parallel laws).
So overall the resistance in the circuit is 1/2.75 ? which is wrong :s

edit:
Whoops made an error with my calculations, I understand it now, thanks emmanuel )
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#6
(Original post by Joinedup)
You've got to break the circuit down into bits you can handle.

What are really easy to handle are circuits containing nothing but resistors in series and nothing but resistors in parallel.

first the two resistors on the right hand side (3R and 6R)... those are a pair of resistors in parallel which can be replaced with a single resistor using the rule for parallel resistors. Reffective = (1/R1 + 1/R2)-1

now the circuit looks like this...

and you've got 3 resistors in series that you can replace with one...

Attachment 531351531355

which you know how to solve as 2 resistors in parallel
Sorry I think the second attachment you provided didnt upload properly? I cant view anything it says the link is invalid
0
#7
(Original post by Joinedup)
You've got to break the circuit down into bits you can handle.

What are really easy to handle are circuits containing nothing but resistors in series and nothing but resistors in parallel.

first the two resistors on the right hand side (3R and 6R)... those are a pair of resistors in parallel which can be replaced with a single resistor using the rule for parallel resistors. Reffective = (1/R1 + 1/R2)-1

now the circuit looks like this...

and you've got 3 resistors in series that you can replace with one...

Attachment 531351531355

which you know how to solve as 2 resistors in parallel
Thank you so much!! I just drew it out and got the answer
Never thought of approaching it from right to left combining the 3 ohm and 6 ohm resistor ahh
Thank you again!
0
3 years ago
#8
(Original post by HasanRaza1)
Sorry I think the second attachment you provided didnt upload properly? I cant view anything it says the link is invalid
Yeah TSR mungs it up when I attach more than one pic... keep thinking they must have fixed it already.

I think the diagram by Eimmanuel probably makes it clearer why you can proceed in this order.
0
3 years ago
#9
(Original post by Joinedup)
You've got to break the circuit down into bits you can handle.

What are really easy to handle are circuits containing nothing but resistors in series and nothing but resistors in parallel.

first the two resistors on the right hand side (3R and 6R)... those are a pair of resistors in parallel which can be replaced with a single resistor using the rule for parallel resistors. Reffective = (1/R1 + 1/R2)-1

now the circuit looks like this...

and you've got 3 resistors in series that you can replace with one...

Attachment 531351531355

which you know how to solve as 2 resistors in parallel
How did you simply in the last step, didnt quite understand that step
0
3 years ago
#10
(Original post by metrize)
How did you simply in the last step, didnt quite understand that step
The steps are:

1) Look for resistors obviously in parallel with each other.

The 3R and 6R are obvious and they can be replaced with a singles resistor of value 2R.

2) Any more parallel? No. Then look for series resistors.

Yes. Sum these and replace with a single resistor. The series resistors are R + 2R + R = 4R.

This is placed in parallel with the existing 4R.

3) Look for parallel resistors.

4R || 4R = 2R.

Circuit cannot be reduced any further. Answer = 2R.
1
3 years ago
#11
(Original post by metrize)
How did you simply in the last step, didnt quite understand that step
Well there are 1R,2R and 1R in series... which can be replaced with 4R

and that just leaves you with 4R and 4R in parallel.
1
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