IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: T.Q on April 27, 2010, 09:47:20 am
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Three identical resistors are connected across a potential difference V so that one of them is in parallel with the other two which are connected in series. The power dissipated through the first one, compared to the power dissipated by each of the other two, is approximately
A the same
B half as much
C twice as much
D four times as much
which one is the answer & explain why ?
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Three identical resistors are connected across a potential difference V so that one of them is in parallel with the other two which are connected in series. The power dissipated through the first one, compared to the power dissipated by each of the other two, is approximately
A the same
B half as much
C twice as much
D four times as much
explain why ?
The answer is D..four times as much..
here's the deal:
Voltage for the lone resistor= V..and voltage for both of the resistors connected in parallel is also V..right?
now the volatage for each of the resistors in series = 0.5V..right?
P=V2 /R
therefore, for lone resistor P=V2 /R
for each of the resistors in series: P= (0.5V)2 /R= 0.25V2 /R
1/0.25=4.
got that? ::)
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ya i got it
thanx :)
+rep
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Any time :)
and Thanks 4 the +rep. :D
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A.F- I told you..you are mind-blasting! :)
+ rep
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Oh :-[ lol
Thanks Nid! :D
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anyone has the ms for physics AS cie 2002 nov
plzzzzzzzzzzz rili urgent
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This here is da link (http://www.xtremepapers.net/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w02_ms_2.pdf)
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This here is da link (http://www.xtremepapers.net/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w02_ms_2.pdf)
thx but wat i rili need is the paper 1 ms :SSS
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The examiner report also includes the paper 1 answers (http://www.xtremepapers.net/CIE/index.php?dir=International%20A%20And%20AS%20Level/9702%20-%20Physics/&file=9702_w02_er.pdf)
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in may06 p1 of phy. cie, why did we split the area inta a trapezium and a triangle instead of just considering the area of the whole triangle????
cud sum1 also xplain Q9. of the same year??????
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Q9) ohk a spring is oscillating up and down... to begin the oscillation you first stretch the string to the lowest point, once released from there it rises to the mean position(at which it has max velocity) A...from A it moves up to the extreme top which is B(velocity is 0). following that, it moves from B back to mean position(now labelled C) and then to the bottom most point which is D(where v is 0 again)
Got the motion?
So, lowest point is D
What no is the trapezium q?
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its Q22
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Q22) it can be called a triangle only upto 10 mm. Line's not straight after that.
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but its written in the Que's to treat the whole line as straight!!!
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but its written in the Que's to treat the whole line as straight!!!
treating XY to be straight only for the calculation of the area of the trapezium...not otherwise (if u extend the rest of the line it wouldn't be close to XY)
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ok... Thank You
can u help me in Q9 of may04
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s= v2 sin 2theta / g tan theta=1.25/10
10= v2 sin (2X7.125)/ 9.8
v2= 10/ 0.025=398.125
v = root of 398.125= 20
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ok... Thank You
can u help me in Q9 of may04
ANOTHER WAY
uy=0
so , s=0.5gt^2
t= (1.25/0.5g)^0.5
t= 0.51 s
v= d/t
v=10/0.51
v= 19.8 ms-1
so the answer is D
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thanx sooo much nid and T.Q
+rep
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u guys could u please help me out in this paper:
paper 1 2002 nov.
Q15
q17
q22
q28
q35
q36
please explain in details thxxx aloot
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u guys could u please help me out in this paper:
paper 1 2002 nov.
Q15
q17
q22
q28
q35
q36
please explain in details thxxx aloot
that's a lot of them...
will take some time
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Let me do the first half.
15: A. Because the diagonal is longer than both the other forces. Thus, it is the hypotenuse. And, the forces are all 90 degress from each other. It is like the end of a box.
17: D. There is no acceleration, so no change in kinetic energy and there for D.
22: A. Solid and liquids have the same densities and the same molecular spacing. The answer is not b, because that would imply that solids have a desnity 1000 times greater than liquids.
28: B. Increasing wavelength means that the chances of interaction decreases, and thus they are further apart.
Nid can do the rest. g2g
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Q15) A obv resultant vector x and y components
Q17) D no change in velocity, therefore no change in k.e
Q22) Density and molecular spacing in solids and liquids is approx the same...hence A
Q28) lambda= ax/ D where a is the distance between the slits, x the distance between the fringes and D is the distance between the slit and screen. to increase separation that is x..
A-> if f is increased means lambda is decreased as v=flambda.....so this is out(cuz lamda and x are directly proportional)
B->lamda increased ...then x will increase...so B
C->decreasing D will reduce x (directly proportional)
D-> increasing a will reduce x (inversely proportional
Q35) R increases with length... so pointer will be moved to Y. But this will cause I=0...V=IR so V=0
Q36) in the parallel wire with the 5ohm resistors, 2V is split equally, so potential between source and X= 2/3 source and Y= 2X2/3=4/3
difference is 2/3 which is A
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thanks for offering help dude...but im done... ;)
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thanks for offering help dude...but im done... ;)
U did get a head start though.
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cud sum1 xplain Q5 of may03??????????
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Q15) A obv resultant vector x and y components
Q17) D no change in velocity, therefore no change in k.e
Q22) Density and molecular spacing in solids and liquids is approx the same...hence A
Q28) lambda= ax/ D where a is the distance between the slits, x the distance between the fringes and D is the distance between the slit and screen. to increase separation that is x..
A-> if f is increased means lambda is decreased as v=flambda.....so this is out(cuz lamda and x are directly proportional)
B->lamda increased ...then x will increase...so B
C->decreasing D will reduce x (directly proportional)
D-> increasing a will reduce x (inversely proportional
Q35) R increases with length... so pointer will be moved to Y. But this will cause I=0...V=IR so V=0
Q36) in the parallel wire with the 5ohm resistors, 2V is split equally, so potential between source and X= 2/3 source and Y= 2X2/3=4/3
difference is 2/3 which is A
for Q36> why do u consider the resistance of X and Y from the source, why not the 3 resistors directly between X and Y ( 2 from the first path and 1 from the second)??????
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huh...how 2 and 1.??
please explain what you are trying to say
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thankyou so muchhhhhhhhhhhhhhh nid and dude 321
rili appreciate it
+ rep
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huh...how 2 and 1.??
please explain what you are trying to say
i meant why dont we consider the 2 resistors to the left of X( on the upper path) and the 1 resistor to the left of Y( the one on the lower path) ???
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for q 36 it is not the restance between X and Y
you have to find the voltage at each point - there is 4V accross each resistor
it goes 0 4 8 12
so if wires are connected at diff poiint they will be at diff voltages and there will be a potential difference