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Unformatted text preview: EE 101 Handout # 27 Professor Abbas El Gamal March # 15, 2003 Homework #8 Solutions 1. (a) During standby mode, v s charges the capacitor through R th up to v s . The time constant for charge-up is (10kΩ)(20 μ F) = 0 . 2sec, so charging is essentially com- plete after a second or two. (Actually, we never asked anything about that.) When the switch is switched to defib, all of the energy stored in the capacitor is (eventually) delivered to the patient’s chest. Hence the energy stored in the capacitor after chargeup, i.e. , Cv 2 s / 2, must equal 100J. Solving, we get v s = 3162V. That’s an impressive voltage, and certainly not one you’d like across your chest (unless your heart has stopped). Another (more complicated) way to solve this problem is to solve for the voltage v chest ( t ), and then integrate the power dissipated, to get the total energy. You get the same answer, of course. But it’s easier (and maybe more elegant) to just note that the stored energy of the capacitor all ends up in dissipated in the patient’s chest. (b) When switched to defib mode, the charged capacitor discharges through the pa- tient’s chest ( R chest ). Let’s assume the switch is thrown at time t = 0 to make our expressions simpler. Since the capacitor is initially charged to 3162V and the time constant of the circuit is 20 μ F · 500Ω = 0 . 01sec. the voltage across R chest is v chest = 3162 e- t/ . 01 . The power through the patient’s chest is therefore p chest ( t ) = v 2 chest /R chest = 20000 e- 200 t . The time T it takes for 90J to be dissipated in R chest can be found by solving for T in the integral 90 = Z T p chest ( t ) dt = Z T 20000 e- 200 t dt which yields T = 11 . 5msec. Another way to find T is to find the time when 10J is left stored in the capacitor, since then the other 90J must have been dissipated in the patient’s chest. This leads to the equation 10 = ( C/ 2) 3162 e- T / . 01 2 (which has, of course, the same solution, T = 11 . 5msec). 1 (c) The maximum power delivered to the patient is just the maximum of p chest ( t ) (found above) for t ≥ 0. The maximum occurs at t = 0, and is p max = 20kW. That’s not a small power — it’s about thirty times the power supplied to a typical microwave oven — so you can see why these early defibrillators often burned patients (in addition to saving their lives). (d) For an RC circuit, all of the energy initially stored in the capacitor is dissipated in the resistor, no matter what its value. Thus when R chest = 1kΩ, 100J is dissipated, the same as when R chest was 500Ω!...
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This note was uploaded on 04/06/2010 for the course EE 102 taught by Professor Pingli during the Fall '08 term at Shanghai Jiao Tong University.
- Fall '08