ent FULL SCREEN | | PRINTER VERSION | BACK NEXT Chapter 20, Problem 103 Your ans

Question

ent FULL SCREEN | | PRINTER VERSION | BACK NEXT Chapter 20, Problem 103 Your answer is partially correct. Try again. In a heart pacemaker, a pulse is delivered to the heart 81 times per minute. The capacitor that controls this pulsing rate discharges through a resistance of 1.6 x 106a-one pulse is delivered every time the fully charged capacitor loses 64.0% of its original charge. What is the capacitance of the capacitor? Number T.0000004729 the tolerance is +/-2% SHOW HINT LINK TO TEXT Question Attempts: Unlimited SAVE FOR LATER SOBMETAI SUBMIT ANSWER

Explanation / Answer

pulse rate = 81 pulses/minute

Therefore, time for one pulse = 60 / 81 = 0.7407 sec

After losing 64% charge, remaining charge is 100 - 64 = 36%

q/qo = 0.36

We know from capacitance equation

q/qo = e-t/RC

Taking ln on both sides, we get

ln(q/qo) = -t/RC

solving for C, we get

ln (0.36) = -0.7407 / 1.6e6*C

-1.02165 = -0.7407 / 1.6e6*C

C = 4.53e-7 F

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