Question

Pacemaker Batteries Pacemakers designed for long-term use commonly employ a lithium-iodine battery capable of supplying 0.42 A?h of charge.

A) How many coulombs of charge can such a battery supply?

deltaQ=_____kC?

B) If the average current produced by the pacemaker is 5.6?A, what is the expected lifetime of the device?

deltaT=_____y?

Answer 1

(3600 s)(A)-(3600 s)(C/s) A-h 3600 C

=>Ans to part A=3.6*0.42 kC=1.512kC

0.42=5.6*10^-6*x hours

=>x=75000 hours=8.5616 years

Answer 2

A) Battery capable of suppling charge = 0.42 A.h

= 0.42*60*60

= 3600kC

B) Average current produced = 5.6*10^^-6

Ampheres per hr = 5.6*10^^-6 / 0.42

= 13.33 *10^-6 A/hr

No. of years (delta T) = 13.33 *10^-6 /24*365

= 0.00152207 *10^-6 y

  

Answer 3

A) deltaQ = total charge that can be supplied

Q = i x t [i=current, t=time]

take all units in SI

0.42 A?h = 0.42 A * 3600 s = 1512 C = 1.512 kC

B) Q = i t

from part A, we calculated total coulomb of charge supplied by the battery = 1.512 kC

given, average current produced by the pacemaker is 5.6?A

thus, i= 5.6 ?A

so, t = Q / i = 1512 C /(5.6 * 10^-6) = 27*10⁷ seconds

$=\frac{27*10^7}{3600 * 24*365} years$

= 8.56 years