# [Physics] Electricity QuestionWatch

#1
Electricity has always been my weakness in A-level physics. Was doing a specimen paper and this was the only question I did really bad on:

I got 3.1 right (I3 = I1 + I2) but can't do the two questions after that.

For 3.2 I tried working out how much p.d. it is getting from V1 which would be 10 I think, then how much it is getting from V2 which would be (30/(10+30)) * 12 so 8. Thus a total of 18V. However the answer is actually 10V.

No idea what to do for 3.3.

Any help would be appreciated, thanks.
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1 year ago
#2
(Original post by DarkEnergy)
I got 3.1 right (I3 = I1 + I2) but can't do the two questions after that.

For 3.2 I tried working out how much p.d. it is getting from V1 which would be 10 I think, then how much it is getting from V2 which would be (30/(10+30)) * 12 so 8. Thus a total of 18V. However the answer is actually 10V.
For 3.1, the answer is 10V, i.e. . You add currents, not voltages.
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1 year ago
#3
(Original post by DarkEnergy)
No idea what to do for 3.3.
You now know the voltage on each side of . What's the voltage drop across it? Use that to calculate .
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#4
(Original post by RogerOxon)
For 3.1, the answer is 10V, i.e. . You add currents, not voltages.
Ah thanks, I get that now.

(Original post by RogerOxon)
You now know the voltage on each side of . What's the voltage drop across it? Use that to calculate .
Oh I see, thank you very much. So I'm guessing it's 2V/10 Ohms
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1 year ago
#5
(Original post by DarkEnergy)
Electricity has always been my weakness in A-level physics. Was doing a specimen paper and this was the only question I did really bad on:

I got 3.1 right (I3 = I1 + I2) but can't do the two questions after that.

For 3.2 I tried working out how much p.d. it is getting from V1 which would be 10 I think, then how much it is getting from V2 which would be (30/(10+30)) * 12 so 8. Thus a total of 18V. However the answer is actually 10V.

No idea what to do for 3.3.

Any help would be appreciated, thanks.
Hi there

For part two, you are using the second law which is 'the sum of the EMFs around a loop = the sum of the potential drops around a loop'. This means that this circuit can be considered as two different loops.

First loop consists of V1, R1 and R3 where the sum of the potential differences across R1 and R3 is equal to the EMF of V1.

The second loop consists of V2, R2 and R3 where the sum of the potential differences across R2 and R3 is equal to the EMF of V2.

For part two, you only need to consider the first loop. If R1 is adjusted to 0 ohms, then there will be no potential difference across is as I x 0 = 0V
Since you know that :
pd across R1 + pd across R3 =10
0 + pd across R3 =10V
Therefore the potential difference across R3 must be 10V.

For the third part consider loop by loop and you should get the answer I hope this helps !

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2
1 year ago
#6
(Original post by RogerOxon)
For 3.1, the answer is 10V, i.e. . You add currents, not voltages.
To expand on this - will drive what ever current is required to keep its voltage at 10V. In this case, it will drive some current to ensure that .

Voltage is like water pressure in your plumbing - it's the differences that drive current (flow) through the resistances. There will be fast transients as the current changes to equalise the voltage (pressure) drops.
1
1 year ago
#7
(Original post by DarkEnergy)
Oh I see, thank you very much. So I'm guessing it's 2V/10 Ohms
Yes, 0.2A.
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#8
(Original post by leopard923)
-
(Original post by RogerOxon)
-
Thank you both very much for the help
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1 year ago
#9
(Original post by RogerOxon)
You now know the voltage on each side of . What's the voltage drop across it? Use that to calculate .
Why do we use voltage drop and not the sum of 10V and 12V = 22V ?
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1 year ago
#10
(Original post by Jas1947)
Why do we use voltage drop and not the sum of 10V and 12V = 22V ?
Voltage is also called Potential Difference, or Electro-Motive Force - it's not something that adds-up. Going back to the plumbing analogy, you take pressure differences across components. If two pipes with the same water pressure join, then you add the currents, not the water pressures.
0
1 year ago
#11
(Original post by RogerOxon)
Voltage is also called Potential Difference, or Electro-Motive Force - it's not something that adds-up. Going back to the plumbing analogy, you take pressure differences across components. If two pipes with the same water pressure join, then you add the currents, not the water pressures.
Thank you

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0
1 month ago
#12
If i2=0.2A. What is i1?
0
1 month ago
#13
(Original post by cosmin0406)
If i2=0.2A. What is i1?

I3R3=10 from part (ii)
hence I3=10/30 = 0.33A

I2R2=2
hence I2=2/10 = 0.20A

From part (i) (KCL),
I3=I1+I2
hence I1=I3-I2=0.33-0.20=0.13A
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