IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: krtcobain82 on May 20, 2011, 03:25:09 am

Title: A-Level Physics paper 2
Post by: krtcobain82 on May 20, 2011, 03:25:09 am
Guys I need explanation for this question.

May/June 08 Paper 2

Q. 6 a) and
Q. 6 b)

http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_s08_qp_2.pdf


Title: Re: A-Level Physics paper 2
Post by: EMO123 on May 20, 2011, 07:59:07 am
6 (a) Use the formula P=IV
        so make I subject of the formula I=P/V
        substitute the values I=(1.5*1000)/240
        :. I=6.25
        now use V=IR
        Make R subject of the formula R=V/I
        Substitute the values R=240/6.25
        :. R=38.4 ohms

Title: Re: A-Level Physics paper 2
Post by: EMO123 on May 20, 2011, 08:03:38 am
6 (b) 1st is open so power supply so that is zero
        2nd from the first the power is supplied so it will move to s2 but s3 is closed so it will not move from s3
              so only of s2 which is 1.5kW so for second it is 1.5kW
        3rd the power is supplied from s2 and s3 so both are added 1.5kW+1.5kW=3kW


i dont understand the 4th and the 5th anybody else who can tell about them plz you may tell
Title: Re: A-Level Physics paper 2
Post by: Ghost Of Highbury on May 20, 2011, 08:43:08 am
Quote from: EMO123 on May 20, 2011, 08:03:38 am
6 (b) 1st is open so power supply so that is zero
        2nd from the first the power is supplied so it will move to s2 but s3 is closed so it will not move from s3
              so only of s2 which is 1.5kW so for second it is 1.5kW
        3rd the power is supplied from s2 and s3 so both are added 1.5kW+1.5kW=3kW


i dont understand the 4th and the 5th anybody else who can tell about them plz you may tell

4th - Current flows through A and B (not the part of the junction with S2 as it has infinite resistance)
        R = 38.4ohms, V one heater receives is 120V , P =  V^2/R = 120^2/38.4 = 375W
        Thus, for two heaters = 375*2 = 750W or 0.75kW
5th - Current flows through all heaters. Power in A and B as calculated previously = 0.75kW
       Power in C = V^2/R = 240^2/38.4 = 1.5kW
Total - 1.5 + 0.75 = 2.25kW
Title: Re: A-Level Physics paper 2
Post by: EMO123 on May 20, 2011, 08:48:19 am
Quote from: Ghost Of Highbury~ on May 20, 2011, 08:43:08 am
4th - Current flows through A and B (not the part of the junction with S2 as it has infinite resistance)
        R = 38.4ohms, V one heater receives is 120V , P =  V^2/R = 120^2/38.4 = 375W
        Thus, for two heaters = 375*2 = 750W or 0.75kW
5th - Current flows through all heaters. Power in A and B as calculated previously = 0.75kW
       Power in C = V^2/R = 240^2/38.4 = 1.5kW
Total - 1.5 + 0.75 = 2.25kW
thanx