Question

Q11. What charges will flow through section
B of the circuit in the direction shown when
switch S is closed.
​

Answer 1

Explanation:

When switch is opened, 2 and 3μF capacitors are in series. So,

C

eq

=

5

2×3

=

5

6

μF

Hence, charge on each capacitor is

q=CV=

5

6

×90=108μC

When switch S is open, let q

1

and q

2

be charges on the two capacitors as in figure (b). So,

q

1

=2×30=60μC

q

2

=3×60=180μC

Let charge q

B

goes to the upper plate 3μF capacitor and lower plate of 2μF capacitor. Initially, both the plates have charge +q−q=0. Finally, they have charges q

2

−q

1

. So

q

2

−q

1

=q

B

+0

or q

B

=q

2

−q

1

=180−60=120μC

Alternatively:

After closing the switch, charge flowing out of 2μF capacitor is

Δq

1

=108−60=48μC

charge flown inot 3μF capacitor is

Δq

2

=180−108=72μC

So the net charge

Δq=Δq

1

+Δq

2

=48+72=120μC

solution