A 13.0-V battery is connected to a 3.70-LF capacitor. How much energy is stored

Question

A 13.0-V battery is connected to a 3.70-LF capacitor. How much energy is stored in the capacitor? Need Help?ReadIt -13 points SerCP11 16.P.051 A parallel-plate capacitor has capacitance 1.00 F (a) How much energy is stored in the capacitor if it is connected to a 15.00-V battery? (b) If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored? (c) The battery is subsequently reattached to the capacitor, but the plate separation remains as in part (b). How much energy is stored?

Explanation / Answer

1)

V = 13.0 V

C = 3.70 uF

= 3.70*10^-6 F

use:

E = 0.5*C*V^2

= 0.5*(3.70*10^-6)*(13.0)^2

= 3.13*10^-4 J

Answer: 3.13*10^-4 J

2)

a)

V = 15.0 V

C = 1.00 uF

= 1.00*10^-6 F

use:

E = 0.5*C*V^2

= 0.5*(1.00*10^-6)*(15.0)^2

= 1.13*10^-4 J

= 113 uJ

Answer: 113 uJ

b)

Energy stores will not change because of conservation of energy

Answer: 113 uJ

c)

C = esoleneo*A/d

when d doubles, C becomes half

So,

E = 0.5*C*V^2

So,

E becomes half

E = 113/2 = 56.5 uJ

Answer: 56.5 uJ

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