Electric Current — Question 8

Section 26-1 · Physics pages 4166–4167

Question 8 A charge budget limits the duty cycle of intermittent operation An ideal charge reservoir has 2.00 Ah usable capacity. A device draws 5.00 mA at rest and 205 mA during a 3.00 s active interval every 60.0 s. Treat the currents as prescribed. t I Axes and labels specify the model; the sketch is schematic. (a) Find the average current and ideal operating time. (b) Find the largest active fraction that permits at least 100 h of operation.
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Question 8 - Solution A charge budget limits the duty cycle of intermittent operation t I (a) Method and calculation f = 3/60 = .0500, I = 5(1 −f) + 205f = 15.0 mA. t = 2.00 Ah/.0150 A = 133.3 h. (b) Interpretation and further calculation I ≤2.00/100 = .0200 A. .005 + .200f ≤.020 ⇒f ≤.0750. Physical check Ampere-hours measure charge, not energy; voltage and nonideal battery effects are excluded.

Question 8

A charge budget limits the duty cycle of intermittent operation

An ideal charge reservoir has 2.00 Ah usable capacity. A device draws 5.00 mA at rest and 205 mA during a 3.00 s active interval every 60.0 s. Treat the currents as prescribed.

Diagram shown in the original SVG worksheet.

  1. Find the average current and ideal operating time.

  2. Find the largest active fraction that permits at least 100 h of operation.

Question 8 - Solution

A charge budget limits the duty cycle of intermittent operation

Diagram shown in the original SVG worksheet.

(a) Method and calculation

f=3/60=.0500,I¯=5(1−f)+205f=15.0mA.f=3/60=.0500,\quad\overline I=5(1-f)+205f=15.0\,\mathrm{mA}.t=2.00Ah/.0150A=133.3h.t=2.00\,\mathrm{Ah}/.0150\,\mathrm A=133.3\,\mathrm h.

(b) Interpretation and further calculation

I¯≤2.00/100=.0200A.\overline I\le 2.00/100=.0200\,\mathrm A..005+.200f≤.020⇒f≤.0750..005+.200f\le.020\Rightarrow f\le.0750.

Physical check

Ampere-hours measure charge, not energy; voltage and nonideal battery effects are excluded.