Other Single-Loop Circuits — Question 10

Section 27-4 · Physics pages 4390–4391

Question 10 A constant-power load admits two operating currents below a source limit A source with emf E and internal resistance r drives an ideal load that absorbs fixed positive power P. Its terminal voltage is positive and P = V I. Assume a static operating point, without specifying load dynamics. + − E P = constant r Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Find the possible currents and the condition for their existence. (b) For E = 10.0 V, r = 1.00 Ωand P = 16.0 W, find both currents and terminal voltages. State what static equations cannot decide.
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Question 10 - Solution A constant-power load admits two operating currents below a source limit + − E P = constant r (a) Method and calculation V = E −rI, P = I(E −rI) ⇒rI2 −EI + P = 0. I± = E ± √ E2 −4rP 2r , P ≤E2/(4r). (b) Interpretation and further calculation I−= 2.00 A, V−= 8.00 V; I+ = 8.00 A, V+ = 2.00 V. I−V−= I+V+ = 16.0 W. Static equations establish possible states; their stability requires a dynamic model. Physical check No real operating point exists above the discriminant limit in this source model.

Question 10

A constant-power load admits two operating currents below a source limit

A source with emf E and internal resistance r drives an ideal load that absorbs fixed positive power P. Its terminal voltage is positive and P=VIP=VI. Assume a static operating point, without specifying load dynamics.

Diagram shown in the original SVG worksheet.

  1. Find the possible currents and the condition for their existence.

  2. For E=10.0E=10.0 V, r=1.00Ωr=1.00\,\Omega and P=16.0P=16.0 W, find both currents and terminal voltages. State what static equations cannot decide.

Question 10 - Solution

A constant-power load admits two operating currents below a source limit

Diagram shown in the original SVG worksheet.

(a) Method and calculation

V=E−rI,P=I(E−rI)⇒rI2−EI+P=0.V=E-rI,\quad P=I(E-rI)\Rightarrow rI^2-EI+P=0.I±=E±E2−4rP2r,P≤E2/(4r).I_\pm=\frac{E\pm\sqrt{E^2-4rP}}{2r},\quad P\le E^2/(4r).

(b) Interpretation and further calculation

I−=2.00A,V−=8.00V;I+=8.00A,V+=2.00V.I_-=2.00\,\mathrm A,\ V_-=8.00\,\mathrm V;\qquad I_+=8.00\,\mathrm A,\ V_+=2.00\,\mathrm V.I−V−=I+V+=16.0W.I_-V_-=I_+V_+=16.0\,\mathrm W.Static equations establish possible states; their stability requires a dynamic model.

Physical check

No real operating point exists above the discriminant limit in this source model.