A 100 pF capacitor is charged by a 50-volt battery. After removing it, that capacitor is then joined with another capacitor. Hence, the potential drops to 35 volts. Find the capacitance of the other.
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Answered by
1
Answer:
Explanation: Q=CV=C1V1+C2V2
200×100=60(200+C2)
As capacitors are connected in parallel voltage across them will be same. So V1=V2
On simplification, C2=133.33 pf
So capacitors capacitance will be 133.33pfn:
kaushikdey:
why did you did c1v1+c2v2
??
and how did you know that capacitors are joined in parallel
Answered by
3
I think it's the correct ans...
hope it helps :)
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in the last... it has written that the potential drops to 35 volts... which means potential both capacitor drops to 35
thanks dear
:)
as potential has dropped so we can say they are in parallel
ohh yaa dear.... :)
very nice question
okz
I hv taken 30 min for this question XD
ha ha
yaa :)
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