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I am not understanding it and how to do their calculation ?????
reference(M/J-2006-Q-7,O/N-2007-Q-8,O/N-2008-Q-11c)
reference(M/J-2006-Q-7,O/N-2007-Q-8,O/N-2008-Q-11c)

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thankswell it says in the 3rd frame 5v/division so that means that 3div=15v etc. when it says 50Hz ac, use the formula F(frequency)= 1/T(time period) so 1 div is equal to 1/50=.02 seconds, that's all i can answer hope it helped!
(M/J-2006-Q-7,O/N-2007-Q-8,O/N-2008-Q-11c) can u explain me this ?well it says in the 3rd frame 5v/division so that means that 3div=15v etc. when it says 50Hz ac, use the formula F(frequency)= 1/T(time period) so 1 div is equal to 1/50=.02 seconds, that's all i can answer hope it helped!
s'06 Q7;(M/J-2006-Q-7,O/N-2007-Q-8,O/N-2008-Q-11c) can u explain me this ?
thanks very muchs'06 Q7;
(A) the 1st socket of the y input will go near the start of the resistor as it is at a higher voltage and the 2nd socket will go after the resistor as it has 0 PD.
B.i) p.d= 3divx2=6v
ii) 1 and a half boxes
n'07
Q8:
same as before. 1/25= .04s/wave hence 1 wave in terms of time= 4 boxes and 2 boxes max amplitude on both sides of mean position.
n'08 Q11c;
(c) (i) as voltage is 0 as it is at the end of the resistor hence pd=0
ii) it gradually increases and amplitude visible, noticeable displacement increases slowly and steadily and finally at M it is maximum ie 6v
iii) when it was 2v/div before total boxes on tboth sides of line= 3 but when it is 1v/div that means 6 boxes so cro screen is too small
hope it helped!
it would help if you clicked like on my commentsthanks very much
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