(a) Use E = (1/2) C V^2 to solve for C.
(b) Since only 1/4 of the stored energy is delivered, you still have sqrt(3/4) = 86.6% of the original voltage on the plates.
(c) "Time constant" = 5*10^-3 s = R*C
Solve for R
In a portable defibrillator, energy is stored in a capacitor charged to a high potential difference. The capacitor
is discharged through the patient’s body in a short, high-powered pulse.
a) The capacitor in a portable defibrillator is charged to 5000 V and stores 800 J of electrical energy. What is its
capacitance?
b) During a pulse, 200 J of energy is delivered to the patient’s body. What is the potential difference across the plates
of the capacitor after a single pulse?
c) We can model the discharging of the capacitor through the patient’s body as an RC circuit, with the heart acting
as the resistor. If a pulse lasts 5 ms, What is the resistance of the path through the heart?
1 answer
2 answers