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Physics: Post your doubts here!

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Paper 5 > how to make l as subject in terms of the gradient ?
(V0/I0)^2 = R^ 2 + 4π^f^ 2L^2

Ans l= ( Vo^2*gradient/4pi^2 )^1/2
 
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Hey guys can anyone please give me a link to notes for applications of physics for CIE ?? I really need it , thank you very much :)
 
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And try to memorize this questions >> they come between 2/3 yrs in a twisted way !
applications is mostly memorizing not like the others !!
 

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hi, can u plz explain the questions 35 from paper 1 oct nov 2007 as level physics?
papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w07_ms_1.pdf
 
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hi, can u plz explain the questions 35 from paper 1 oct nov 2007 as level physics?
papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20(9702)/9702_w07_ms_1.pdf
In figure 1, the ammeter shows the reading of the current that is passing through the two resistors in the upper parallel wire. Furthermore, the voltmeter shows the difference in voltages between two points. Seeing at the upper connection of the voltmeter, we know that the total potential drop across the upper two resistors would be 6V. Thus in one resistor, 3v would have already been dessipated, and there would be 3V at the upper terminal of the voltage. Same happens in the lower parallel wire. There would be 3V at this terminal too. As the difference between the two readings is zero, the voltmeter gives a reading of zero.


For figure 2, the ammeter once again is showing the current passing through 2 resistors, thus the reading would be same. Furthermore, we can see that the terminals of the voltmeter are once again connected after one resistor in both of the parallel wires. Thus, there would again be no pottential difference between the two points.

P.S just look at terminals of the voltmeter, and do not confuse it with the circuit
 
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Note that we r asked to find I2 i.e. the current across a fixed resistor..
so maximum current will be:- V/R= 12/6 = 2.0 A
and i think that the minimum current value is zero just bcuz if we take the value of the variable resistor less than the value of fixed resistor then all the current will flow through the low resistance loop giving zero reading on the ammeter for the second loop
 
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Hi, I was doing this paper http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf and I got stuck hardcore on question 37. The answer should be C apparently but it makes no logical sense to me. Can someone explain it to me?
the answer is C becuz here we have 2 resistors that will half the value of the current...
Suppose that the value of one resistor is 2 ohms n the voltage supply is 12V
so for first circuit current will be:- 12/2= 6A
while for the second circuit it will be:- 12/4=3A (NOTE: resistors are in series so they will add up to 4)
thats the reason y the current value won't remain the same..so it is incorrect
 
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Hi, I was doing this paper http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w12_qp_12.pdf and I got stuck hardcore on question 37. The answer should be C apparently but it makes no logical sense to me. Can someone explain it to me?

Maybe you didn´t read the question properly? They are asking for which row is NOT correct. Maybe that´s why you dont understand why the answer is C? The answer is C because it´s wrong! - Thats what they are asking for! :D
 
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