A defibrillator passes a brief burst of current through the heart to restore nor

Question

A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a capacitor is charged to 3.3 kV. Paddles are used to make an electric connection to the patient's chest. The paddles are attached to the patient using an electrolytic gel which minimizes the skin contact resistance. As the result the electrical resistance of the patient (from paddle to paddle) is only 232.1 . At the time t=1 ms the current through the patient is 9.3 A.

What is the current through the patient 3.3 ms later?

Explanation / Answer

the growth of current in RC circuit is

I(t) = Io ( 1- e^-t/ RC)

9.3 A = ( 3.3* 10 ^3 V/232.1) ( 1- e^1 * 10 ^-3 s/ 232.1 (C)

9.3/14.21 =  ( 1- e^1 * 10 ^-3 s/ 232.1 (C)

0.65 = ( 1- e^1 * 10 ^-3 s/ 232.1 (C)

e^-1 * 10 ^-3 s/ 232.1 (C)= 0.345

take ln on both sides

-1 * 10 ^-3 s/ 232.1 (C) = ln ( 0.345)

-1 * 10 ^-3 s/ 232.1 ( -1.06)= C

C = 4.05 * 10 ^-6 F

--------------------------------------------------------------

I(t) = Io ( 1- e^-t/ RC)

i(t) = ( 3.3* 10 ^3 V/232.1) ( 1- e^-3.3 * 10 ^-3 s/ 232.1 (4.05 * 10 ^-6 F)

=13.79 A

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