Potential Difference Between Two Points — Question 9

Section 27-5 · Physics pages 4408–4409

Question 9 A prescribed set of voltage readings must satisfy loop consistency Three stationary circuit nodes have claimed readings VB −VA = 5.00 V, VC −VB = −2.00 V and VA −VC = −4.00 V. A single reading contains an additive error; the first two are trusted. A B C +5 V −2 V −4 V claimed Schematic; labels, not drawing lengths, specify the data. (a) Test the closed-path sum and determine the corrected third reading. (b) If a particle of charge q travels from A to C, find its potential-energy change from either trusted path.
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Question 9 - Solution A prescribed set of voltage readings must satisfy loop consistency A B C +5 V −2 V −4 V claimed (a) Method and calculation (VB −VA) + (VC −VB) + (VA −VC) = 5 −2 −4 = −1.00 V̸ = 0. VC −VA = 5 −2 = 3.00 V ⇒(VA −VC)correct = −3.00 V. (b) Interpretation and further calculation ∆UA→C = q(VC −VA) = 3.00q V. q[(VB −VA) + (VC −VB)] = q(5 −2) V = ∆UA→C. Physical check The consistency test concerns electrostatic node potentials; circuits with appreciable time-varying magnetic flux require a broader treatment.

Question 9

A prescribed set of voltage readings must satisfy loop consistency

Three stationary circuit nodes have claimed readings VB−VA=5.00V_B-V_A=5.00 V, VC−VB=−2.00V_C-V_B=-2.00 V and VA−VC=−4.00V_A-V_C=-4.00 V. A single reading contains an additive error; the first two are trusted.

Diagram shown in the original SVG worksheet.

  1. Test the closed-path sum and determine the corrected third reading.

  2. If a particle of charge q travels from A to C, find its potential-energy change from either trusted path.

Question 9 - Solution

A prescribed set of voltage readings must satisfy loop consistency

Diagram shown in the original SVG worksheet.

(a) Method and calculation

(VB−VA)+(VC−VB)+(VA−VC)=5−2−4=−1.00V≠0.(V_B-V_A)+(V_C-V_B)+(V_A-V_C)=5-2-4=-1.00\,\mathrm V\ne0.VC−VA=5−2=3.00V⇒(VA−VC)correct=−3.00V.V_C-V_A=5-2=3.00\,\mathrm V\Rightarrow(V_A-V_C)_{\mathrm{correct}}=-3.00\,\mathrm V.

(b) Interpretation and further calculation

ΔUA→C=q(VC−VA)=3.00qV.\Delta U_{A\to C}=q(V_C-V_A)=3.00q\,\mathrm V.q[(VB−VA)+(VC−VB)]=q(5−2)V=ΔUA→C.q[(V_B-V_A)+(V_C-V_B)]=q(5-2)\,\mathrm V=\Delta U_{A\to C}.

Physical check

The consistency test concerns electrostatic node potentials; circuits with appreciable time-varying magnetic flux require a broader treatment.