Question

The figure shows a 11.8 V battery and four uncharged capacitors of capacitances C1 = 1.16 μF,C2 = 2.31 μF,C3 = 3.26 μF, and C4 = 4.19 μF. If only switch S1 is closed, what is the charge on (a) capacitor 1, (b) capacitor 2, (c) capacitor 3, and (d) capacitor 4? If both switches are closed, what is the charge on (e) capacitor 1, (f) capacitor 2, (g) capacitor 3, and (h) capacitor 4?

Answer 1

given

V = 11.8 volts

C₁ = 1.16 μF,

= 1.16 x 10^-6 F

C₂ = 2.31 μF,

= 2.31 x 10^-6 F

C₃ = 3.26 μF,

= 3.26 x 10^-6 F

and C₄ = 4.19 μF

= 4.19 x 10^-6 F

If only switch S₁ is closed ,

C_n1 = C₁ C₃ / C₁ + C₃

= 1.16 x 3.26 x 10^-12 / ( 1.16 + 3.26 ) x 10^-6

C_n1 = 8.555 x 10^-7 F

Q_n1 = C_n1 V

= 8.555 x 10^-7 x 11.8

Q_n1 = 1 x 10^-5 C

Q₁ = Q₃ = Q_n1 = 1 x 10^-5 C

C_n2 = C₂ C₄ / ( C₂ + C₄ )

= 2.31 x 4.19 x 10^-12 / ( 2.31 + 4.19 ) x 10^-6

C_n2 = 1.489 x 10^-6 F

Q_n2 = C_n2 V

= 1.489 x 10^-6 x 11.8

Q_n2 = 1.757 x 10^-5 C

Q₂ = Q₄ = Q_n2 = 1.757 x 10^-5 C

If both switches are closed,

C_n1 ' = C₁ + C₂

= ( 1.16 + 2.31 ) x 10^-6 = 3.47 x 10^-6 F

C_n2 ' = C₃ + C₄

= ( 3.26 + 4.19 ) x 10^-6 = 7.45 x 10^-6 F

Cn ' = C_n1 ' C_n2 ' / ( C_n1 ' + C_n2 ' )

= 3.47 x 10^-6 x 7.45 x 10^-6 / (3.47 x 10^-6 + 7.45 x 10^-6 )

Cn ' = 2.367 x 10^-6 F

Q_n ' = C_n' x V

= 2.367 x 10^-6 x 11.8

Q_n ' = 2.793 x 10^-5 C

Q₁ ' = Q_n ' = 2.793 x 10^-5 C

V₁ ' = Q₁' / C_n1 '

= 2.793 x 10^-5 / 3.47 x 10^-6 F

V₁ ' = 8.048 volts

Q₁ = C₁ V₁ '

= 1.16 x 10^-6 x 8.048

Q₁ = 9.333 x 10^-6 C

Q₂ = C₂ V₁ '

= 2.31 x 10^-6 x 8.048

Q₂ = 18.59 x 10^-6 C

Q₂ ' = Q_n ' = 2.793 x 10^-5 C

V₂ ' = Q₁' / C_n2 '

= 2.793 x 10^-5 / 7.45 x 10^-6

V₂ ' = 3.748 volts

Q₃ = C₃ V₁ '

= 3.26 x 10^-6 x 3.748

Q₃ = 12.218 x 10^-6 C

Q₄ = C₄ V₁ '

= 4.19 x 10^-6 x 3.748

Q₄ = 15.7 x 10^-6 C