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[QUOTE="justinawe;42980662"]Let the resistance of the single resistor be R1, and the resistance of the parallel combination be R2.

R1=200000 ΩR_1 = 200000 \ \Omega

R2=(1200000+1200000)1=100000 ΩR_2 = \left( \dfrac{1}{200000} + \dfrac{1}{200000} \right)^{-1} = 100000 \ \Omega

Now, there are two ways you could do this.

You could either find the current using:

V=IRV = IR

V=I(R1+R2)[/tex[br][br][tex]12=I(200000+100000)[/tex][br][br][tex]I=4×105 A[/tex][br][br]Then,(V[sub]2[/sub]isthepdacrosstheparallelcombination)[br][br][tex]V2=IR2[/tex][br][br][tex]V2=4×105×100000=4 V[/tex][br][br][br][br]Or,alternatively,youcouldusethefactthat:[br][br][tex]VR[/tex][br][br]Whichmeansthat,[br][br][tex]V1:V2=R1:R2[/tex][br][br][tex]R1:R2=V1:V2=200000:100000=2:1[/tex][br][br]Usingknowledgeofbasicratios,andthefactthatV[sub]1[/sub]+V[sub]2[/sub]=12,youcaneasilyarriveatthecorrectanswerhere.V = I(R_1 + R_2)[/tex[br][br][tex]12 = I(200000+100000)[/tex][br][br][tex]I = 4 \times 10^{-5} \ \mathrm{A}[/tex][br][br]Then, (V[sub]2[/sub] is the pd across the parallel combination)[br][br][tex]V_2 = IR_2[/tex][br][br][tex]V_2 = 4 \times 10^{-5} \times 100000 = 4 \ \mathrm{V}[/tex][br][br][br][br]Or, alternatively, you could use the fact that:[br][br][tex]V \propto R[/tex][br][br]Which means that,[br][br][tex]V_1 : V_2 = R_1 : R_2[/tex][br][br][tex]R_1 : R_2 = V_1 : V_2 = 200000:100000 = 2:1[/tex][br][br]Using knowledge of basic ratios, and the fact that V[sub]1[/sub]+V[sub]2[/sub]=12, you can easily arrive at the correct answer here.

Thank you so much!
Reply 581
Good luck to all!!!
Reply 582
Good luck all! 1 hour 40 minutes to the exam
Good luck to all/
Reply 584
Hope Edexcel do not mess up
Good luck all! Let's kick Physics in the bawlls!
Good luck lads
exam in 1 hr
Hi guys. Thought it was a good paper, until it got to the last question :colondollar: I put resistance would be lower at a lower current because of less heating, so fewer ionic vibrations and less electron-ion collisions. Seems like I was in the minority. Thoughts?
(edited 10 years ago)
Reply 589
was the circuit on the second last question in series or parallel?
Original post by GCSE-help
Hi guys. Thought it was a good paper, until it got to the last question :colondollar: I put resistance would be lower at a lower current because of less heating, so fewer ionic vibrations and less electron-ion collisions. Seems like I was in the minority. Thoughts?


Original post by mny355
was the circuit on the second last question in series or parallel?


I put resistance would be lower for the same reasons you did and in series

Posted from TSR Mobile
Original post by mny355
was the circuit on the second last question in series or parallel?


Series, because if one bulb went out the other one did too.
Reply 592
What did everyone get for the multiple choice questions?
I put it in series too. Okay starting from MCQs what did u put for first two ? I put violin instrument and mgh/VIt
Original post by frannyw
What did everyone get for the multiple choice questions?


I got mgh/VIT for the efficiency, even though some said it was 0.5mv^2/VIT.

I also put superposition for the final MC but apparently it was polarisation. Silly me for not reading the q :frown:
Reply 595
Original post by GCSE-help
I got mgh/VIT for the efficiency, even though some said it was 0.5mv^2/VIT.

I also put superposition for the final MC but apparently it was polarisation. Silly me for not reading the q :frown:


I did 0.5mv^2/VIT, but assume I'm wrong, and I put superposition too :/
For last Mcq I put polarization. and for one with angle of incidence and angle or refraction I did 90- whatsoever the angle with horizontal. and angle of refraction was smaller value.
Original post by jollygood
For last Mcq I put polarization. and for one with angle of incidence and angle or refraction I did 90- whatsoever the angle with horizontal. and angle of refraction was smaller value.


Yeah you're right on both counts but could you please remind me about the polarisation question? Are you sure it wasn't superposition? I didn't read the question and hastily ticked superposition after seeing the image.

Also what did you put for the last question?
Okay and there wasn't any other hard MCQ apart from these two three. now coming to the questions answers first three were fairly okay. For the resistance one in which the real value was double the calculated value I said that by applying force the wire becomes longer as well as thinner so cross sectional area also decreases and for this reason the ressitance is twice the calculated value. what did u guys put?
Original post by GCSE-help
Yeah you're right on both counts but could you please remind me about the polarisation question? Are you sure it wasn't superposition? I didn't read the question and hastily ticked superposition after seeing the image.

Also what did you put for the last question?


Yeah sure the last mcq was about the two waves passing through thin film of oil and producing many different colours or something like that but it was same diagram as in one of past paper. and it asked which phenomenon isnot involved in producing different colours? it was polarization.

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