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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: conceptual qn....
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shubham_sachdeva (1793)

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Q1.a capacitor of capacity C is charged to a potential of V...another capacitor of capacitance 2C & potential 2V. now they are connected in parallel in such a way that +ve plateof 1st capacitor is connected with -ve plate of 2nd capacitor. write expression for initial , final & loss in energy...
 
Q2.if there are two capacitors A and B  connected in series such that there is a key in /w them..i hope dia. is clear...now qn. satrts..if q charge is given to capacitor A. then when that key is closed then what amount of charge will flow through it after a long time......??
 

padhna likhna chad de mitra , nakal te rakhi aas , chak chaddar te so ja bhagta , Rabb karega tennu paas.....

PLZ NEVER EVER RATE ME FOR MY ANSWERS , IF U WANT TO COMPLIMENT ME THEN JUST BEAT UR HEAD & SAY
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rooney (861)

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its 2 in the night. Will type tomorrow morning. Good night.

http://14-69-8.blogspot.com
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shubham_sachdeva (1793)

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Olaaa!! Perrrfect answer. 309  [433 rates]

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u will be welcomed...

padhna likhna chad de mitra , nakal te rakhi aas , chak chaddar te so ja bhagta , Rabb karega tennu paas.....

PLZ NEVER EVER RATE ME FOR MY ANSWERS , IF U WANT TO COMPLIMENT ME THEN JUST BEAT UR HEAD & SAY
"YEH KIS PAGAL NE MERA QN. SOLVE KER DIYA"

I AM SERIOUS!!!!
EVEN SERIOUS ^ INFINITY
PLZZZZZZ NEVER RATE ME....HOPE U WILL UNDERSTAND....

Shubham


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abhijeet_0201 (756)

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1) since +ve plate of 1st capacitor is connected to - ve plate of second capacitor
they are in series
eq capacitanceC'=C1C2/C1+C2

=5C/3
now since they are in series
eq pot diff V'=V1+V2
V'=2V+V=3V

initial energy=( C1V1^2)/2 + (C2V2^2)/2
(8+1)/2 CV^2=4.5CV^2

final energy= (C'V'^2)/2
=3CV^2

so loss in energy=(4.5-3)CV^2
=1.5CV^2


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