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Physics / Chapter 27: Circuits / Section 27-6

Multiloop Circuits — Question 9

Section 27-6 · Physics pages 4428–4429

Question 9 Only independent loop equations add new information Three branches join the same top and bottom nodes. The left branch has an upward emf E1 and resistor r1; the right has upward E2 and r2; the middle has resistor R. Let I1 and I2 flow upward in the outer branches and IL downward through R. + − E1 r1 + − E2 r2 R V Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Write three independent equations sufficient to determine the three branch currents. (b) Show algebraically that the outer-loop equation follows from the two smaller-loop equations and cannot replace the junction rule.
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Question 9 - Solution Only independent loop equations add new information + − E1 r1 + − E2 r2 R V (a) Method and calculation I1 + I2 −IL = 0. E1 −r1I1 −RIL = 0, E2 −r2I2 −RIL = 0. (b) Interpretation and further calculation (E1 −r1I1 −RIL) −(E2 −r2I2 −RIL) = 0. E1 −E2 −r1I1 + r2I2 = 0. This relation is redundant; it contains no additional independent constraint. Physical check Drawing more loops does not necessarily provide more independent equations.

Question 9

Only independent loop equations add new information

Three branches join the same top and bottom nodes. The left branch has an upward emf E1 and resistor r1; the right has upward E2 and r2; the middle has resistor R. Let I1 and I2 flow upward in the outer branches and IL downward through R.

Diagram shown in the original SVG worksheet.

  1. Write three independent equations sufficient to determine the three branch currents.

  2. Show algebraically that the outer-loop equation follows from the two smaller-loop equations and cannot replace the junction rule.

Question 9 - Solution

Only independent loop equations add new information

Diagram shown in the original SVG worksheet.

(a) Method and calculation

I1+I2−IL=0.I_1+I_2-I_L=0.E1−r1I1−RIL=0,E2−r2I2−RIL=0.E_1-r_1I_1-RI_L=0,\quad E_2-r_2I_2-RI_L=0.

(b) Interpretation and further calculation

(E1−r1I1−RIL)−(E2−r2I2−RIL)=0.(E_1-r_1I_1-RI_L)-(E_2-r_2I_2-RI_L)=0.E1−E2−r1I1+r2I2=0.E_1-E_2-r_1I_1+r_2I_2=0.This relation is redundant; it contains no additional independent constraint.

Physical check

Drawing more loops does not necessarily provide more independent equations.

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