1due to electromagnetic induction the charges become 5 each
A 4f capacitor and 6f capacitor are connected in parallel across a 660 v parallel source. The charged capacitors are now disconnected from the line and from each other with terminals of unlike signs together.find the final charges on each and voltage across each other.
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8 Answers
1hello asish
first they are connected in parallel so voltage across each will be same so charge on each will be 2640μC (on 4μF) and 3960μC (on 6μF)..
finally when they are reconnected ..conserving the charge we get 1320 μC charge
potential will become same
so let q1 be the chrge on 4μF and q2 be the chrge on 6μF
q1+q2= 1320
and q1/4 = q2/6
q2=792μC
q1=528μC
3shriya there is a slight error .............. its q1/4 + q2/6 = 0 ......... see they are connected with unlike signs together ...............................[tellme if mah concepts are messed up ]
1i also agree....it must be....q1/4+q2/6=0..........
hey asish.....wht d ans...?????
1asish ihnk u have missed a word....."are disconnected and from each other with...." in this line...
106oops yeah they r disconnected from each other and reconnected with unlike plates together. (negative plate of one with positive plate of other) .. but i think