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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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