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Physics: Post your doubts here!

biscuitbiscuit

biscuitbiscuit

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usama321 said:
You have got to see the current now, not voltage. The voltage across the two components would be different as they are in series now. in series the current is the same. Therefore, you will have to see the voltages for the same amount of curent
Click to expand...
In previous post you said that voltage across 'C' would be 5v out of 7, how did you get that?
 
usama321

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biscuitbiscuit said:
In previous post you said that voltage across 'C' would be 5v out of 7, how did you get that?
Click to expand...
Nah, i was saying that just to eliminate any factor of doubt. Just to clarify, the current in the circuit can not increase more than around 2.5, because after that, the combined p.d across the two components would become more than the emf... Thus before the value of around 2.5 ma, all the values of the pd across C are more than the resistor, leading to more power being dissipated across it.
 
biscuitbiscuit

biscuitbiscuit

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usama321 said:
Nah, i was saying that just to eliminate any factor of doubt. Just to clarify, the current in the circuit can not increase more than around 2.5, because after that, the combined p.d across the two components would become more than the emf... Thus before the value of around 2.5 ma, all the values of the pd across C are more than the resistor, leading to more power being dissipated across it.
Click to expand...
Thank you sooo much! :)
 
Thought blocker

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Mohammed salik said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_13.pdf
Q11 and 26 Plz Thanks :)
Thought blocker
Click to expand...
Q26 is solved.
sara kamal said:
for 26 answer is B
since 2 x distance between 2 nodes=wavelength
so as in 1m 3nodes are there ,the distance between 2 nodes =1/3=o.333
hence wavelength=2 x o.333=0.666m
so f=330/o.666=495Hz
Click to expand...

Q11 :¬

P initial = 2mu - mu =mu
And for momentum to be conserved, P ini = P final :)
So check it for all question..
Option A)
-((2mu)/3)+(5mu/3) = 3mu/3 = mu so momentum is conserved Pi = Pf
Option b)
-((2mu)/6)+(2mu/3) = 1/3mu = Momentum is NOT conserved Pi =/(not equal) pf :)
So answer is option B :)
 
biscuitbiscuit

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Thought blocker said:
Then please solve it for me :oops:
Click to expand...
a) Power is directly proportional to V^2, Resistance is constant.
Therefore % reduction is, (230)^2-(220)^2/(230)^2*100 = 8.5%

b)(i) ammeter reading would be zero since p.d. is zero when C is at A
(ii) at B the p.d. would be 1.5 and according to the formula, I=V/R, I= 1.5/0.5= 0.3A

c(i) You have to plot these points on the graph and draw a line of best fit.
(ii) Read the current when l= 50cm
(iii) V= IR, therefore V= 0.12*5= 0.6V
:)
 
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biscuitbiscuit said:
a) Power is directly proportional to V^2, Resistance is constant.
Therefore % reduction is, (230)^2-(220)^2/(230)^2*100 = 8.5%

b)(i) ammeter reading would be zero since p.d. is zero when C is at A
(ii) at B the p.d. would be 1.5 and according to the formula, I=V/R, I= 1.5/0.5= 0.3A

c(i) You have to plot these points on the graph and draw a line of best fit.
(ii) Read the current when l= 50cm
(iii) V= IR, therefore V= 0.12*5= 0.6V
:)
Click to expand...
I dont get a part.
 
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