QUESTION If the voltage across the capacitor were doubled, the energy stored wou

Question

QUESTION If the voltage across the capacitor were doubled, the energy stored would be multiplied by: PRACTICE IT Use the worked example above to help you solve this problem. A fully charged defibrillator contains 1.25 KJ of energy stored in a 1.10 x 10-4 F capacitor. In a discharge through a patient, 6.15x102J of electrical energy are delivered in 2.60 ms. (a) Find the voltage needed to store 1.25 k] in the unit. (b) What average power is delivered to the patient? EXERCISE HINTS: GETTING STARTED I I'M STUCK! (a) Find the energy contained in a 2.00 x 10-5 F parallel-plate capacitor if it holds 1.75 x 10-3 C of charge. (b) What's the voltage between the plates? (c) What new voltage will result in a doubling of the stored energy'? Need Help? L-Read it-

Explanation / Answer

Given

fully charged defibtillator contains energy U = 1.25 kJ

capacitor is C = 1.10*10^-4 F

which is discahrging through a patient 615 J of energy in 2.60 ms

a) voltage needed to store 1.25 kJ is

U = 0.5*C*V^2

1250 = 0.5*1.10*10^-4*v^2

solving for V

v = 4767.3 V

b) average power delivered to the patient is P = U/t = 615/(2.60*10^-3) W = 236538.46 W

ExERCISE

a) C = 2*10^-5 F,q = 1.75*10^-3 C

U1 = q^2/2*c

U1 = (1.75*10^-3)^2/(2*2*10^-5) J

U1 = 0.0765625 J

b) potential or voltage between the plates is  

V1 = Q/C

V1 = (1.75*10^-3)/(2*10^-5)V

V1 = 87.5 V

c) to double the energy u2 = 2*U1

U = 0.5*C*v^2

U2/U1 = v2^2/v1^2

v2^2 = v1^2*U2/U1

v2 = sqrt(v1^2*U2/U1)

v2 = sqrt(87.5^2*2) V

V2 = 123.74 V

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