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# DC electricity question watch

1. http://www.scribd.com/mobile/doc/53418189?width=768

For question 5, D is the answer. But WHY is D the answer? I'm terrible at DC electricity in unit 2, so please explain it fully so I understand it.

Thanks.
2. Here is an image that I have created showing you how to work out the answer, hope it helps!

I used 7V just as an example!

P.S. I hated electricity!

3. (Original post by x0diak)
Here is an image that I have created showing you how to work out the answer, hope it helps!

I used 7V just as an example!

P.S. I hated electricity!

Say if N takes 1V, then there is 6V left over. So each lamp in parallel has 6V each while N takes 1V.

With M removed, both N and L take 3.5V each.

So N has a higher voltage and L has a lower voltage. So in the exam question, why can't answer C be correct?
Thanks
4. Hmmm I'm not too sure for that one.

All I can say is that they would probably be expecting you to recognise them as all being the same lamp - therefore having a maximum voltage around half the voltage of the supply.

The point you make is a good one and I am not sure of what the answer to it would be.
5. (Original post by x0diak)
Hmmm I'm not too sure for that one.

All I can say is that they would probably be expecting you to recognise them as all being the same lamp - therefore having a maximum voltage around half the voltage of the supply.

The point you make is a good one and I am not sure of what the answer to it would be.
Okay, thanks for the help anyways. Btw, I hate electricity as well. It's so hard to understand the content and visualise it compared to other modules in Physics
6. Couldn't agree more!
7. This is a bit trickier than it looks as you need to know several things before it can be answered.

The key fact is learning that the current flowing into a junction is the same as that leaving the junction.

So the current going into lamp N is the same as the added currents going through L and M. i.e. another way of saying the current in lamp L or M is half that of lamp N.

The total resistance of the circuit is RN + (RL in parallel with RM). i.e. if all the resistances are the same, the total circuit resistance is R + 0.5R = 1.5R.

The brightness of each lamp is determined by how much power is consumed in each.

P=IxV (triangle) but you also know that V = I/R (ohms law triangle).

So doing a quick substitution for V in the first equation gives:

P=IxV which is equal to P=Ix(I/R) = I2R.

Hence the power output (brightness) of lamp N is I2R
But the power output of each lamp L and M is (I2R/2)/2 = I2R/4 because the current is split between the lamps. So each lamp L and M will only be 1/4 as bright as lamp N.

But now if lamp M blows it has gone open circuit, the total resistance of the circuit is now R + R = 2R. So the total current in the circuit has gone down because the voltage is the same but the resistance has gone up.

Which means that the brightness of lamp N must go down as well.

But, all of the current that was previously split between the two lamps L and M now goes through L only. Hence it must gets brighter.

Lots to follow but hope it helps.
8. (Original post by uberteknik)
This is a bit trickier than it looks as you need to know several things before it can be answered.

The key fact is learning that the current flowing into a junction is the same as that leaving the junction.

So the current going into lamp N is the same as the added currents going through L and M. i.e. another way of saying the current in lamp L or M is half that of lamp N.

The total resistance of the circuit is RN + (RL in parallel with RM). i.e. if all the resistances are the same, the total circuit resistance is R + 0.5R = 1.5R.

The brightness of each lamp is determined by how much power is consumed in each.

P=IxV (triangle) but you also know that V = I/R (ohms law triangle).

So doing a quick substitution for V in the first equation gives:

P=IxV which is equal to P=Ix(I/R) = I2R.

Hence the power output (brightness) of lamp N is I2R
But the power output of each lamp L and M is (I2R/2)/2 = I2R/4 because the current is split between the lamps. So each lamp L and M will only be 1/4 as bright as lamp N.

But now if lamp M blows it has gone open circuit, the total resistance of the circuit is now R + R = 2R. So the total current in the circuit has gone down because the voltage is the same but the resistance has gone up.

Which means that the brightness of lamp N must go down as well.

But, all of the current that was previously split between the two lamps L and M now goes through L only. Hence it must gets brighter.

Lots to follow but hope it helps.
I can't understand the bold part where you somehow get I^2R/4. Could you please explain that part fully?
Thanks

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