Question

. A defibrillator sends 12 A of current through a patient’s torso for 3.0x10^ - 3 seconds.

A) If this pulse of current deposited 300 J of energy in the patient during that time, what was the voltage during that time? (Think about power.)

(Answer: 8.3x10 3 V)

B) What is the resistance of the current’s path through the patient?

(Answer: 690 ohms )

The correct answers are given. I just don't know how to get there. Thank you in advance!

Answer 1

Solution) (A)Current i = 12 A

Time t = 3×10^(-3) s

Energy E = 300 J

Voltage V = ?

We have Power P = (V)(i)

V = (P/i)

But Power P = (Energy/time ) = (E/t)

P = ((300)/(3×10^(-3))

P = 10^(5) W

V = (P/i) = ((10^(5))/(12))

V = 8.3×10^(3) V

Therefore voltage V = 8.3×10^(3) V

(B) Resistance R = ?

We have voltage V = (i)(R)

R = (V/i)

R = ((8.3×10^(3))/(12))

R = 691 ohms

Approximately Resistance R = 690 ohms