can someone answer http://www.freeexampapers.us/IGCSE/Physics/CIE/2008%20Jun/0625_s08_qp_3.pdf
NUMBER 8 A THAAANKS AND PHYSICS PAPER 1 WAS AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAMAZING LOTSA TRICKS THO
Ahhh I hope ours is just as good! Congrats By the way
for a you draw them all in parallel and you put a switch near the battery. then 3 others on each line where the lamps are.
for b) P=IV so I= P/V so 100/200 is 0.5 A. I might be wrong though check and tell me
for b)ii) I(current)= Q(charge in coulombs)/t(time in secs) 1 min=60 secs.
Q=I x t so 0.5A x 60 is 30C
for c)i) you add the total power of the first 3 lamps minus the total power of the second three lamps
ii) you take the value that you got for for the reduction in power (c)i)) over 360 secs (1x60x60)
Check the correct answer and tell me whether my answers are even remotely correct