Other Single-Loop Circuits — Question 9

Section 27-4 · Physics pages 4388–4389

Question 9 Nearly balanced sources can leave current direction unresolved by tolerances Two opposing sources have hard emf bounds E1 = 12.0 ± .2 V and E2 = 11.9 ± .3 V. Total loop resistance is exactly 2.00 ohm. Define positive current in the E1-driven direction. + − E1 R + − E2 I Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Find nominal current and its exact interval. (b) Find the allowed resistor-heating interval and explain why the minimum cannot be obtained simply by squaring the two current endpoints.
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Question 9 - Solution Nearly balanced sources can leave current direction unresolved by tolerances + − E1 R + − E2 I (a) Method and calculation I = (E1 −E2)/R, I0 = (12.0 −11.9)/2 = .0500 A. I ∈[(11.8 −12.2)/2, (12.2 −11.6)/2] = [−.200, .300] A. (b) Interpretation and further calculation 0 ∈[Imin, Imax] ⇒Pmin = 0. Pmax = R max(I2 min, I2 max) = 2(.300)2 = .180 W. Physical check The uncertainty interval permits either current direction and includes exact cancellation.

Question 9

Nearly balanced sources can leave current direction unresolved by tolerances

Two opposing sources have hard emf bounds E1=12.0±.2E_1=12.0\pm.2 V and E2=11.9±.3E_2=11.9\pm.3 V. Total loop resistance is exactly 2.00 ohm. Define positive current in the E1-driven direction.

Diagram shown in the original SVG worksheet.

  1. Find nominal current and its exact interval.

  2. Find the allowed resistor-heating interval and explain why the minimum cannot be obtained simply by squaring the two current endpoints.

Question 9 - Solution

Nearly balanced sources can leave current direction unresolved by tolerances

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I=(E1−E2)/R,I0=(12.0−11.9)/2=.0500A.I=(E_1-E_2)/R,\quad I_0=(12.0-11.9)/2=.0500\,\mathrm A.I∈[(11.8−12.2)/2,(12.2−11.6)/2]=[−.200,.300]A.I\in[(11.8-12.2)/2,(12.2-11.6)/2]=[-.200,.300]\,\mathrm A.

(b) Interpretation and further calculation

0∈[Imin,Imax]⇒Pmin=0.0\in[I_{\min},I_{\max}]\Rightarrow P_{\min}=0.Pmax=Rmax⁡(Imin⁡2,Imax⁡2)=2(.300)2=.180W.P_{\max}=R\max(I_{\min}^2,I_{\max}^2)=2(.300)^2=.180\,\mathrm W.

Physical check

The uncertainty interval permits either current direction and includes exact cancellation.