JIPMER 2019 Morning Papers Online Practice

(3)
When switch is open:

In series capacitors, charge is same
⇒ CV = constant
⇒

VA

VB

=

CB

CA

(C ‌A = 3µF, C ‌B = 6µF)
⇒

VA

VB

=

6

3

=

2

1

also V‌A + V‌B = 18V
⇒ V‌A = 12V & V‌B = 6V
V ‌1 = 18 – V‌A = 6V
In series resistance, (R ‌C = 3Ω, R ‌D = 6Ω)

VC

VD

=

RC

RD

VC

VD

=

3

6

=

1

2

VC+VD = 18V
⇒ V‌C = 6V & V‌D = 12D
so V ‌2 = 18 – V‌C = 12V
so, charge, q‌A = 3 × 12µC = 36µC
q‌B = 6 × 6 µC = 36µC
On closing switch, V‌1 will become equal to V‌2

At steady state,

VC

VD

=

3

6

VC+VD = 18V
V‌C = 6V, V‌D = 12V
Final charge, on A
q ‌A = 6 × 3 = 18µC
q‌B = 6 × 12 = 72µC