"Pumping" Charges — Question 8

Section 27-1 · Physics pages 4326–4327

Question 8 Finite energy-conversion power limits a charge pump operating at fixed voltage A source converts mechanical power Pm = 80.0 W into electrical output with efficiency eta=.750. It maintains terminal voltage V=24.0 V within its operating range. low V high V positive charge non-electrostatic pump Schematic; labels, not drawing lengths, specify the data. (a) Find its maximum steady current and charge delivered in 30.0 s. (b) If the demanded current is 3.00 A at the same voltage, find required mechanical power and decide whether the specified source can sustain it.
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Question 8 - Solution Finite energy-conversion power limits a charge pump operating at fixed voltage low V high V positive charge non-electrostatic pump (a) Method and calculation Pe = ηPm = 60.0 W, Imax = Pe/V = 2.50 A. Q = Imaxt = (2.50)(30.0) = 75.0 C. (b) Interpretation and further calculation Pm,required = V I/η = (24.0)(3.00)/.750 = 96.0 W. 96.0 > 80.0 ⇒the specified voltage and current cannot both be sustained. Physical check An ideal voltage description must still be restricted to the physical energy-conversion capacity of a real source.

Question 8

Finite energy-conversion power limits a charge pump operating at fixed voltage

A source converts mechanical power Pm=80.0P_m=80.0 W into electrical output with efficiency eta=.750. It maintains terminal voltage V=24.0 V within its operating range.

Diagram shown in the original SVG worksheet.

  1. Find its maximum steady current and charge delivered in 30.0 s.

  2. If the demanded current is 3.00 A at the same voltage, find required mechanical power and decide whether the specified source can sustain it.

Question 8 - Solution

Finite energy-conversion power limits a charge pump operating at fixed voltage

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Pe=ηPm=60.0W,Imax=Pe/V=2.50A.P_e=\eta P_m=60.0\,\mathrm W,\quad I_{\max}=P_e/V=2.50\,\mathrm A.Q=Imaxt=(2.50)(30.0)=75.0C.Q=I_{\max}t=(2.50)(30.0)=75.0\,\mathrm C.

(b) Interpretation and further calculation

Pm,required=VI/η=(24.0)(3.00)/.750=96.0W.P_{m,\mathrm{required}}=VI/\eta=(24.0)(3.00)/.750=96.0\,\mathrm W.96.0>80.0⇒the specified voltage and current cannot both be sustained.96.0>80.0\Rightarrow\text{the specified voltage and current cannot both be sustained}.

Physical check

An ideal voltage description must still be restricted to the physical energy-conversion capacity of a real source.