GOAL Apply energy and power concepts to a capacitor. PROBLEM A fully charged def
ID: 1577360 • Letter: G
Question
GOAL Apply energy and power concepts to a capacitor. PROBLEM A fully charged defibrillator contains 1.20 kJ of energy stored in a 1.10x 104 F capacitor. In a discharge through a patient, 6.00 102 J of electrical energy are delivered in 2.50 ms. (a) Find the voltage needed to store 1.20 k) in the unit. (b) What average power is delivered to the patient? STRATEGY Because we know the energy stored and the capacitance, we can use the equation below to find the required voltage in part (a). For part (b), dividing the energy delivered by the time gives the average power. Energy stored·Low-laur- SOLUTION (a) Find the voltage needed to store 1.20 kJ in the unit. Solve the equation above for . tnervy stored·law 21.200 1.10 x 10*F 467 x10 (b) What average power is delivered to the patient? Divide the energy delivered by the 2.50 20-3 -2.40 x 10s w LEARN MORE REMARKS The power delivered by a draining capacitor isn't constant, as we'll find in the study of RC circuits. For that reason, we were able to find only an average power. Capacitors are necessary in defibrillators because they can deliver energy far more quickly than batteries. Batteries provide current through relatively slow chemical reactions, whereas capacitors can quickly release charge that has already been produced and stored QUESTION If the voltage across the capacitor were doubled, the energy stored would be multiplied by:Explanation / Answer
Practice it
Given
U = 1.28 kJ = 1.28*10^3 J
C = 1.10*10^-4 F
dU = 6.12*10^2 J
dt = 2.6 ms = 2.6*10^-3 s
a)
use, U = (1/2)*C*V^2
==> V = sqrt(2*U/C)
= sqrt(2*1.28*10^3/(1.1*10^-4))
= 4824 V
b) Average power delivered, P_avg = dU/dt
= 6.12*10^2/(2.6*10^-3)
= 2.35*10^5 W
Excericise
a) U = Q^2/(2*C)
= (1.75*10^-3)^2/(2*2.8*10^-5)
= 0.0547 J
b) V = Q/C
= 1.75*10^-3/(2.8*10^-5)
= 62.5 V
c) V' = sqrt(2)*V
= sqrt(2)*62.5
= 88.4 V
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