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Physics / Chapter 27: Circuits / Section 27-5

Potential Difference Between Two Points — Question 1

Section 27-5 · Physics pages 4392–4393

Question 1 A mixed branch requires signed potential changes rather than unsigned drops A branch carries 2.00 A from A to D. In order from A to D it contains a 3.00 ohm resistor, an ideal 9.00 V source crossed from negative to positive, and a 1.00 ohm resistor. The rest of the circuit maintains the stated current. 3 Ω + − 9 V 1 Ω A B C D 2 A Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Find VD-VA and the potentials at the two internal nodes if VA=0. (b) Trace the branch backward to verify VA-VD and identify which source supplies energy to the branch current.
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Question 1 - Solution A mixed branch requires signed potential changes rather than unsigned drops 3 Ω + − 9 V 1 Ω A B C D 2 A (a) Method and calculation VB −VA = −2(3) = −6.00 V, VC −VB = +9.00 V. VB = −6.00 V, VC = 3.00 V, VD = 3 −2(1) = 1.00 V. (b) Interpretation and further calculation VA −VD = +2(1) −9 + 2(3) = −1.00 V. Psource = 9(2) = 18.0 W > 0. Crossing its emf in the current direction raises charge energy. Physical check A branch can have a net potential rise while also dissipating heat in its resistors.

Question 1

A mixed branch requires signed potential changes rather than unsigned drops

A branch carries 2.00 A from A to D. In order from A to D it contains a 3.00 ohm resistor, an ideal 9.00 V source crossed from negative to positive, and a 1.00 ohm resistor. The rest of the circuit maintains the stated current.

Diagram shown in the original SVG worksheet.

  1. Find VD-VA and the potentials at the two internal nodes if VA=0.

  2. Trace the branch backward to verify VA-VD and identify which source supplies energy to the branch current.

Question 1 - Solution

A mixed branch requires signed potential changes rather than unsigned drops

Diagram shown in the original SVG worksheet.

(a) Method and calculation

VB−VA=−2(3)=−6.00V,VC−VB=+9.00V.V_B-V_A=-2(3)=-6.00\,\mathrm V,\quad V_C-V_B=+9.00\,\mathrm V.VB=−6.00V,VC=3.00V,VD=3−2(1)=1.00V.V_B=-6.00\,\mathrm V,\ V_C=3.00\,\mathrm V,\ V_D=3-2(1)=1.00\,\mathrm V.

(b) Interpretation and further calculation

VA−VD=+2(1)−9+2(3)=−1.00V.V_A-V_D=+2(1)-9+2(3)=-1.00\,\mathrm V.Psource=9(2)=18.0W>0.P_{\mathrm{source}}=9(2)=18.0\,\mathrm W>0.Crossing its emf in the current direction raises charge energy.

Physical check

A branch can have a net potential rise while also dissipating heat in its resistors.

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