Question

1. (a) Find the charge stored on each capacitor in Figure 18.24 (C₁ = 30.0 µF, C₂ = 4.50 µF) when a 1.51 V battery is connected to the combination.
C₁
___C?
C₂
___C?
0.300 µF capacitor
___C?

(b) What energy is stored in each capactitor?
C₁
___J?
C₂
___ J?
0.300 µF capacitor
____J ?

2.(a) What voltage must be applied to the 10.0 µF capacitor of a heart defibrillator to store 450 J in it?
____ kV ?
(b) Find the amount of stored charge.
___ C ?

Answer 1

1. Given that ::

capacitor, C₁ = 30 x 10^-6 F

capacitor, C₂ = 4.5 x 10^-6 F

voltage of the battery, V = 1.51 V

(a) charge stored on each capacitor C₁ is given as :

Q₁ = C₁ V { eq.1 }

inserting the values in above eq.

Q₁ = (30 x 10^-6 F) (1.51 V)

Q₁ = 45.3 x 10^-6 C

charge stored on each capacitor C₂ is given as :

Q₂ = C₂ V { eq.2 }

inserting the values in eq.2,

Q₂ = (4.5 x 10^-6 F) (1.51 V)

Q₂ = 6.79 x 10^-6 C

charge stored on each capacitor 0.300 F is given as :

Q₃ = C₃ V { eq3 }

inserting the values in eq.3,

Q₃ = (0.3 x 10^-6 F) (1.51 V)

Q₃ = 0.45 x 10^-6 C

(b) energy is stored in each capactitor is given as ::

For capacitor C₁,

E₁ = (0.5) C₁ V²                                        { eq.4 }

E₁ = (0.5) (30 x 10^-6 F) (1.51 V)²

E₁ = 34.2 x 10^-6 J

For capacitor C₂,

E₂ = (0.5) C₂ V                                    { eq.5 }

E₂ = (0.5) (4.5 x 10^-6 F) (1.51 V)²

E₂ = 5.13 x 10^-6 J

For capacitor 0.3 F,

E₃ = (0.5) C₃ V²                                                { eq.6 }

E₃ = (0.5) (0.3 x 10^-6 F) (1.51 V)²

E₃ = 0.342 x 10^-6 J

2. (a) given that :

capacitance of a capacitor, C = 10 x 10^-6 F

energy stored, E = 450 J

voltage must be applied which given as ::

E = 1/2 C V²                                                      { eq.7 }

inserting the values in eq.

(450 J) = (0.5) (10 x 10^-6 F) V²

V² = (450 J) / (5 x 10^-6)

V = 90 x 10^-6

V = 0.00948 V

V = 9.48 kV

(b) the amount of stored charge is given as ::

Q = C V                                                  { eq.8 }

inserting the values in eq.

Q = (10 x 10^-6 F) (0.00948 V)

Q = 0.0948 x 10^-6 C

Q = 94.8 x 10^-9 C