Question

A potential difference of 330 V is applied to a series connection of two capacitors, of capacitance C₁ = 1.60 μF and capacitance C₂ = 8.20 μF.

(a) What is the charge q₁ on capacitor 1?
C

(b) What is the potential difference V₁ across capacitor 1?
V

(c) What is the charge q₂ on capacitor 2?
C

(d) What is the potential difference V₂ on capacitor 2?
V

The charged capacitors are then disconnected from each other and from the battery. Then the capacitors are reconnected with plates of the same signs wired together (the battery is not used).

(e) What now is q₁?
C

(f) What is V₁?
V

(g) What is q₂?
C

(h) What is V₂?
V

Suppose, instead, the capacitors charged in part (a) are reconnected with plates of opposite signs wired together.

(i) What now is q₁?
C

(j) What is V₁?
V

(k) What is q₂?
C

(l) What is V₂?
V

Answer 1

Given V= 330V

C₁ = 1.6μF

C₂ = 8.2μF

Total capacitance C = 1/C₁+ 1/C₂ = (1/1.6+1/8.2)x 10^-6 = 625000 +121951.23 =0.747μF

Now

a)       General equation for Charge, Q =CV

Q₁ = C₁ V = 1.6x10^-6 x 330 = 5.28x10^-4 C or 528μC

b)       Voltage across capacitor C₁ is

V₁ =Q/C₁ ; where Q is the total charge Q = C V = 0.747x10^-6 x 330 =2.465x10^-4C

Now V₁ = Q/C₁ = (2.465x10^-4 )/(1.6x10^-6 )=154.069V

c)       Q₂ = C₂ V = (8.2x10^-6) x (330V) = 2.706 x10^-3C or 2.706mC

d)       Voltage across capacitor C₂ is

V₂ = Q/ C₂ = (2.465x10^-4) / (8.2x10^-6) =30.061V