- #1
TheGreenMarin
- 18
- 0
Homework Statement
The problem involves a battery of emf 10V [with no internal resistance] and a combination of resistors. Here is the diagram provided:
A picture of the diagram can be found here: http://gyazo.com/7791710d0645de38cfab59f0ac4740ec
(Couldn't get the image feature to work)
The previous questions asked for:
- Total resistance, which I found to be 5Ω, which is correct
- Next part says show that each resistor is 3Ω, which I did correctly
The part that I am stuck on is calculating the current through resistor Y.
Homework Equations
R=V/I is the only equation I need to use
The Attempt at a Solution
The question can be found here (its question 7): http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA1-W-QP-JUN11.PDF
The answers can be found here: http://store.aqa.org.uk/qual/gce/pdf/AQA-PHYA1-W-MS-JUN11.PDF
Since the pd across Y needs to be found before I try to figure out the current, I simply used V = IR, this gave me 6V across Y (2A x 3Ω). So then I simply did I = V/R which gave me an answer of 2A (6V / 3Ω).
My answer was clearly wrong as current splits in parallel so it cannot be 2A which was stated in the question as the reading of the ammeter, which was in series.
I couldn't figure it out, so I checked the mark scheme and it said I had to do:
10V (emf) - (2Ax3Ω) which gives you 4V across Y and then you get 1.3A across Y.
My question is, why do I have to take away the voltage across Y from the emf? Surely that would just give me the remaining emf in the circuit, not the pd across Y.