Question

Question 2.25: Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage across each capacitor.

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-90

Answer 1

see figure, here $C_2\:and\:C_3$ are in series , so, $1/C'=1/C_2+1/C_3\implies C"=\frac{200\times200}{200+200}=100pF$
also $C'\:and\:C_1$ are in parallel,
so, C" = 100pF + 100pF = 200pF

$\because\:\:\:C_4\:and\:C"$are in series , so net capacitance of the network is $1/C_{eq}=1/C"+1/C_4$
Ceq = 200 × 100/(200 + 100) pF
Ceq = 200/3 = 66.7 pF

net charge stored on the combination is
Q = Ceq.V = 66.7 × 10^-12 × 300
= 2 × 10^-8 C

As C" and C₄ are in series , so Q" = Q₄ = Q = 2 × 10^-8C [ because in series charges on each are same ]
and hence, V" = Q"/V" = 2× 10^-8/200× 10^-12
V" = 100V
and V₄ = Q₄/C₄ = 2 × 10^-8/100 × 10^-12
V₄ = 200V

it is clear that C₁ and C' are in parallel.
so, V₁ = V' = V" = 100V [ because in parallel potential difference are same ]
and hence, Q₁ = C₁ V₁ = 100 × 10^-12 × 100
Q₁ = 1 × 10^-8 C
similarly, Q' = C'V' = 100 × 10^-12 × 100
Q' = 1 × 10^-8 C

because, C₂ and C₃ are in parallel.
so, Q₂ = Q₃ = Q' = 1 × 10^-8 C
and hence, V₂ = Q₂/C₂ = 1 × 10^-8C/200 × 10^-12
V₂ = 50V
and V₃ = Q₃/C₃ = 1 × 10^-8/200 × 10^-12
V₃ = 50V