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Q.

Three capacitors of capacitances, 2 pF, 3 pF and 4 pF are connected in parallel. 

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a

If 3pF is shorted effective capacity is 2pF.

b

The total capacitance of combination is equal to  1213pF

c

Total capacitance is equal to 9 pF.

d

The total charge is equal 900 pC when connected to 100 V supply.

answer is B, C.

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Detailed Solution

V=100q=CV
So  2×1012×100=Q1
 3×1012×100=Q24×1012×100=Q3
Total charge Q1+Q2+Q3
900×1012=900pC
C-effective =C1+C2+C3  
2+3+4=9pF
 

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