Guest

In the given circuit, a charge of +80 µC is given to the upper plate of the 4 µF capacitor. Then in the steady state, the charge on the upper plate of the 3 µF capacitor is +80 µC 4 µF 3 µF 2 µF (A) +32 µC (B) +40 µC (C) +48 µC (D) +80 µC its an iitjee 2012 paper 2 question … icant get how to apply laws here

In the given circuit, a charge of +80 µC is given to the upper plate of the 4 µF capacitor. Then in the steady state, the charge on the upper plate of the 3 µF capacitor is +80 µC 4 µF 3 µF 2 µF (A) +32 µC (B) +40 µC (C) +48 µC (D) +80 µC 
 
its an iitjee 2012 paper 2 question … icant get how to apply laws here

Grade:12

1 Answers

Vikas TU
14149 Points
3 years ago
The total charge on plate is 80μC. If qb and qc charges on plates B and C then
 
 qb + qc =80....(1)
As the capacitors B and C are in parallel so potential across both are equal .
i.e,
qb/Cb = qc/Cc   
 qc = 48μC
Now for sign of charge: as lower plate of C is connected to ground so upper plate of C should be positive.
Thus, charge on upper plate of 3μF is +48μC

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free