[Physics] Electricity Question

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.

Reply 1

RogerOxon

Original post by DarkEnergy

For 3.1, the answer is 10V, i.e.

V_1

. You add currents, not voltages.

Reply 2

RogerOxon

Original post by DarkEnergy

No idea what to do for 3.3.

You now know the voltage on each side of

R_2

. What's the voltage drop across it? Use that to calculate

I_2

Reply 3

DarkEnergy

Original post by RogerOxon

For 3.1, the answer is 10V, i.e.

V_1

. You add currents, not voltages.

Ah thanks, I get that now.

Original post by RogerOxon

You now know the voltage on each side of

R_2

. What's the voltage drop across it? Use that to calculate

I_2

Oh I see, thank you very much. So I'm guessing it's 2V/10 Ohms

(edited 6 years ago)

Reply 4

leopard923

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:

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 :smile:

I hope this helps !

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Reply 5

RogerOxon

Original post by RogerOxon

For 3.1, the answer is 10V, i.e.

V_1

. You add currents, not voltages.

To expand on this -

V_1

will drive what ever current is required to keep its voltage at 10V. In this case, it will drive some current to ensure that

V_1=I_3R_3

.

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.

Reply 6

RogerOxon

Original post by DarkEnergy

Oh I see, thank you very much. So I'm guessing it's 2V/10 Ohms

Yes, 0.2A.

Reply 7

DarkEnergy

Original post by leopard923

Original post by RogerOxon

Thank you both very much for the help

Reply 8

Minnie16

Original post by RogerOxon

You now know the voltage on each side of

R_2

. What's the voltage drop across it? Use that to calculate

I_2

Why do we use voltage drop and not the sum of 10V and 12V = 22V ?

Reply 9

RogerOxon

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.

Reply 10

Minnie16

Original post by RogerOxon

Thank you

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Reply 11

cosmin0406

If i2=0.2A. What is i1?

Reply 12

username3249896

Original post by cosmin0406

If i2=0.2A. What is i1?

This is an old thread

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

Reply 13

eznot

Because the voltage across R3 is 10V does that mean the voltage across R2 would be 2V so that the voltage in the right-hand loop adds up to 12V???