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 seen in Figure 21.38.

(a) What is the time constant if an 8.00-μF capacitor is used and the path resistance through her body is 1.00 x 10³ Ω? 

(b) If the initial voltage is 10.0 kV, how long does it take to decline to 5.00 x 10² V?

Strategy

Since the resistance and capacitance are given, it is straightforward to multiply them to give the time constant asked for in part (a). To find the time for the voltage to decline to 5.00 x 10² V, we repeatedly multiply the initial voltage by 0.368 until a voltage less than or equal to 5.00 x 10² V is obtained. Each multiplication corresponds to a time of τ seconds.

Data given in Figure 21.38

Transcribed Image Text:

R ww I Switch (a) V = V₁ e-/RC (discharging). +q V Von capacitor V 0.368 V 0 0 T = RC 2r (b) 3r 4r