Edexcel Physics Unit 2 - Waves

Where the chap uses two voltmeters?

Yeah, but the part where you have to calculate the value of R. Also, when doing that part, do you assume that one of the voltmeters is replaced with an ammeter or do you keep it as two voltmeters? Because if you keep it as two voltmeters, then surely no current will get through so you won't be able to tell what the resistance is.

current will get through. But the reason you need an ammeter is to measure the current. And the ammeter should replace V1. V2 is in the correct place, as it measures the potential difference accross the component.

Regarding V1, it is 10M(ohm). Which is a huge number, in total it uses up all the energy and there is none left for the component with 100 ohm.

And how do you calculate the value of R?

the voltage going in V2 = 3V.
The resistance of the component = 100 (ohm)
Work out the current by V = IR

I = V/R
I = 3V/100 (ohm)
I = 0.03 Ampere

Now you know the EMF = 9V and the voltage accross the component = 3V

the voltage accross R = 9V - 3V = 6V
The current is the same accross the series circuit. Hence it is 0.03 Ampere (as calculated before)

Now use V = IR to work out the value of R.

R = V/I
R = 6V/0.03A
R = 200 (ohm)

Its simple and easy as that.

And how do you calculate the value of R?

I found a copy of the answer sheet online and it says the answer is 10Mohms.

http://www.fizix.info/securedocs/alevel/gce08/examzone_markschemes/unit2topic4_examzone_ms.pdf

If you need a username and password, user is student, password is bosco.

We're calculating the young's modulus of copper wire for our practical. We're doing it at the moment as well. I don't do any other sciences with edexcel though. What are you doing for your practical?