A heart defibrillator is used to resuscitate an accident victim by discharging a capacitor through the trunk of her body. A simplified version of the circuit is shown . (a) What is the time constant if an 8.00-μF capacitor is used and the path resistance through her body is 1.00×103 Ω ? (b) If the initial voltage is 10.0 kV, how long does it take to declineto 5.00×102 V ? V = Ve/RC(discharging).(21.80)capacitorVonV -0.368V.0 T= RC 2r3r4rSwitch(a)(b)Figure 21.39 (a) Closing the switch discharges the capacitor C through the resistor R . Mutual repulsion of like charges on each plate drives thecurrent. (b) A graph of voltage across the capacitor versus time, with V = Vo at t = 0. The voltage decreases exponentially, falling a fixedfraction of the way to zero in each subsequent time constant t.

(a) Write the expression for the time constant for the discharging circuit.

Answered: V = Ve/RC (discharging). (21.80)…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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A heart defibrillator is used to resuscitate an accident victim by discharging a capacitor through the trunk of her body. A simplified version of the circuit is shown . (a) What is the time constant if an 8.00-μF capacitor is used and the path resistance through her body is 1.00×10³ Ω ? (b) If the initial voltage is 10.0 kV, how long does it take to decline
to 5.00×10² V ?

V = Ve/RC
(discharging).
(21.80)
capacitor
Von
V -
0.368V.
0 T= RC 2r
3r
4r
Switch
(a)
(b)
Figure 21.39 (a) Closing the switch discharges the capacitor C through the resistor R . Mutual repulsion of like charges on each plate drives the
current. (b) A graph of voltage across the capacitor versus time, with V = Vo at t = 0. The voltage decreases exponentially, falling a fixed
fraction of the way to zero in each subsequent time constant t.

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