JEE Advanced Model Paper 6 with solutions for online practice

When switch S is open, the equivalent capacitance is
C =

C1C2

C1+C2

=

2×4

(2+4)

=

4

3

µF
Charge on the R.H.S. plate of C1 and upper plate of C2 is

Q = CE =

4

3

µF × 3V = 4µC
When switch S is closed, the potential difference across C1 is zero, because the batteries inopposition. Therefore, charge on the R.H.S. plate of C1 = 0 . If the charge flowing through Bis q, then
Q + q = 0 ⇒ q = -Q = -4µC
Now the charge on the upper plate of C2 = C2E = 4µF × 3V = 12µC
which is the chargeflowing through A.