# ELECTRICITY multiple choice Question HELP!Watch

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Thread starter 2 years ago
#1
How do I do this?
0
2 years ago
#2
(Original post by Kraixo)
How do I do this?
- Combine the two 2 Ohm resistors in parallel (the bottom and diagonal ones)
- Combine that in series with the 2 Ohm resistor (left one connected to X)
- Combine that in parallel with the last 2 Ohm resistor (the one between X & Y)

0
2 years ago
#3
(Original post by Kraixo)
How do I do this?
So that you can check, I make it B.
1
2 years ago
#4
(Original post by RogerOxon)
So that you can check, I make it B.
I got 1.2 as well
0
2 years ago
#5
I'm not the best at electricity but I think you can redraw the circuit like so and then work out the parallel ones uning the formula, then add it to the ones in series. Hoping the photo shows up!

However I got 5 ohms like that so I'm probably completely wrong.
0
Thread starter 2 years ago
#6
(Original post by JustJusty)
I'm not the best at electricity but I think you can redraw the circuit like so and then work out the parallel ones uning the formula, then add it to the ones in series. Hoping the photo shows up!

However I got 5 ohms like that so I'm probably completely wrong.
@

RogerOxon

Is the above circuit the same as the one in the question?
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Thread starter 2 years ago
#7
(Original post by RogerOxon)
- Combine the two 2 Ohm resistors in parallel (the bottom and diagonal ones)
- Combine that in series with the 2 Ohm resistor (left one connected to X)
- Combine that in parallel with the last 2 Ohm resistor (the one between X & Y)

I followed what you said and got 1.2 which is the correct answer. However, how do you know which resistor is in parallel and series with what?
0
2 years ago
#8
(Original post by Kraixo)
@

RogerOxon

Is the above circuit the same as the one in the question?
Yes.

(Original post by Kraixo)
I followed what you said and got 1.2 which is the correct answer. However, how do you know which resistor is in parallel and series with what?
Look at where the current can flow. If there's more than one branch that it can take, then those branches are in parallel.

From X, the current can go straight to Y, or down, so we have the top (x-Y) 2 Ohm resistor in parallel with the rest of the circuit. Going down, we go through the 2 Ohm resistor, so that's in series with the rest of this branch. From the bottom left, we can go through the diagonal or bottom, so they're in parallel.
1
2 years ago
#9
I keep on getting 0.8ohms. I add the left and top on cos their in series. i add in parallel the bottom and diagonal.
why are the left and top not in series.
0
2 years ago
#10
isn't the 2 on the left resistor in series with the top resistor... same current through them
0
2 years ago
#11
(Original post by revisionphysics)
isn't the 2 on the left resistor in series with the top resistor... same current through them
No. You have two parallel branches
- the top resistor
- the left one in series with the other two in parallel

Any current from X through the left resistor has to go through either the bottom or diagonal resistor to get to Y.
0
2 years ago
#12
(Original post by Kraixo)
@

RogerOxon
Is the above circuit the same as the one in the question?
Everything is in the same place and connected by the same wires, the wires just have more bends in them.

Why would you calculate the last one in parallel? Apart from the bottom one they're in series, on one loop. :/
0
2 years ago
#13
(Original post by RogerOxon)
No. You have two parallel branches
- the top resistor
- the left one in series with the other two in parallel

Any current from X through the left resistor has to go through either the bottom or diagonal resistor to get to Y.
so we are basically saying that the current at X splits downwards and towards Y so current in them are different and they are therefore in parallel.
0
2 years ago
#14
(Original post by anissabir95)
I keep on getting 0.8ohms. I add the left and top on cos their in series. i add in parallel the bottom and diagonal.
why are the left and top not in series.
Left and top are not in series. Top and (left in series with the other two in parallel) are in parallel.

(Original post by RogerOxon)
- Combine the two 2 Ohm resistors in parallel (the bottom and diagonal ones)
1 Ohm
- Combine that in series with the 2 Ohm resistor (left one connected to X)
3 Ohm

- Combine that in parallel with the last 2 Ohm resistor (the one between X & Y)
2 | 3 Ohm = 6/5 = 1.2 Ohm
0
2 years ago
#15
(Original post by Kraixo)
@

RogerOxon
Is the above circuit the same as the one in the question?
this circuit is drawn wrong.

see this picture for the real circuit...

http://prntscr.com/fjs7f0
0
2 years ago
#16
(Original post by revisionphysics)
this circuit is drawn wrong.

see this picture for the real circuit...

http://prntscr.com/fjs7f0
They're the same.
0
2 years ago
#17
(Original post by RogerOxon)
They're the same.
ye but with this you can see that the left and top resistors are in parallel its
cause with that ciruit i would use

1/R=1/r +1/r +1/r

so 1/R=1/2 +1/2 +1/4 =1.25
and then get R=0.8 which is wrong

but with this you will get R=1.2
0
2 years ago
#18
(Original post by revisionphysics)
ye but with this you can see that the left and top resistors are in parallel its
cause with that ciruit i would use

1/R=1/r +1/r +1/r

so 1/R=1/2 +1/2 +1/4 =1.25
and then get R=0.8 which is wrong

but with this you will get R=1.2
?

They're the same, so you'd do the same calculation.
0
2 years ago
#19
(Original post by RogerOxon)
?

They're the same, so you'd do the same calculation.
ffs... no time to do this...
0
Thread starter 2 years ago
#20
(Original post by RogerOxon)
They're the same.
Bro please explain this too. then answer is 1/4

0
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