Qualification > Sciences
ALL PHYSICS DOUBTS - POST HERE !!
Ghost Of Highbury:
--- Quote from: Ivo on May 26, 2010, 08:14:26 pm ---Sorry, here it is: http://www.freeexampapers.us/IGCSE/Physics/CIE/2009%20Nov/0625_w09_qp_31.pdf
I don't understand how it is worked out.
--- End quote ---
10ai) Step up transformer obviously.
aii) Less current, less electrical energy lost, less heat energy given out
b) P = V*I, I = 55k/22k = 2.5
c) Rate of loss formula, P = I^2R = 2.5^2*3 = 18.75
d) The overall voltage reduced from 22K Volts due to the resistance of the wire, till it reaches the receiving end.
I know this is confusing sh*t, it had appeared in my paper! nov 09! V = 2.5*3 = 7.5
from wiki "For example, an electric space heater may very well have a resistance of ten ohms, and the wires which supply it may have a resistance of 0.2 ohms, about 2% of the total circuit resistance. This means that 2% of the supplied voltage is actually being lost by the wire itself."
e) 22000 - 7.5 -7.5 = 21985 (2 times 7.5 because there are two wires, each voltage dorp = 7.5)
Naruto123:
GUYZ do we have to change da temperature to kelvin or not ??? thx
Vin:
--- Quote from: 33oomar on May 26, 2010, 08:15:48 pm ---guys i need help in q8 May / JUne 2007 Paper 3 Part B
PLEASE
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b) i)
P =I V
I = P/V
=36/1
=3A
ii) V = IR
R = V/I
=12/3
= 4 ohm
iii) 1/R = 1/R1 + 1/R
1/R = 1/4 + 1/4
1/R = 2/4 = 1/2
so, R = inverse of 1/2 = 2 ohm
iv) E = P t
= 36 * 30
=1080 J
Malak:
http://www.freeexampapers.us/IGCSE/Physics/CIE/2009%20Nov/0625_w09_qp_31.pdf
Q11 part c nd so on the rest
thx
Adzel:
--- Quote from: 33oomar on May 26, 2010, 08:15:48 pm ---guys i need help in q8 May / JUne 2007 Paper 3 Part B
PLEASE
--- End quote ---
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