Multiloop Circuits — Question 7

Section 27-6 · Physics pages 4424–4425

Question 7 A voltage source between unknown nodes creates a supernode constraint Nodes A and B connect to ground through 2.00 ohm and 3.00 ohm respectively. An ideal 5.00 V source connects A to B with B positive. An external current driver injects 3.00 A into A and returns through ground. + − 5 V 2 Ω 3 Ω A B 3 A ground Circuit connections and marked polarities define the model; wire lengths are schematic. (a) Use a combined-node current equation and the voltage constraint to find VA and VB. (b) Find the current through the 5.00 V source from A to B and its delivered power.
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Question 7 - Solution A voltage source between unknown nodes creates a supernode constraint + − 5 V 2 Ω 3 Ω A B 3 A ground (a) Method and calculation VA/2 + VB/3 = 3, VB −VA = 5. 3VA + 2(VA + 5) = 18 ⇒VA = 1.60 V, VB = 6.60 V. (b) Interpretation and further calculation IAB = VB/3 = 2.20 A = 3 −VA/2. P5 = 5IAB = 11.0 W delivered. Physical check The unknown current through an ideal voltage source is found from a node balance after the potentials are solved.

Question 7

A voltage source between unknown nodes creates a supernode constraint

Nodes A and B connect to ground through 2.00 ohm and 3.00 ohm respectively. An ideal 5.00 V source connects A to B with B positive. An external current driver injects 3.00 A into A and returns through ground.

Diagram shown in the original SVG worksheet.

  1. Use a combined-node current equation and the voltage constraint to find VA and VB.

  2. Find the current through the 5.00 V source from A to B and its delivered power.

Question 7 - Solution

A voltage source between unknown nodes creates a supernode constraint

Diagram shown in the original SVG worksheet.

(a) Method and calculation

VA/2+VB/3=3,VB−VA=5.V_A/2+V_B/3=3,\quad V_B-V_A=5.3VA+2(VA+5)=18⇒VA=1.60V,VB=6.60V.3V_A+2(V_A+5)=18\Rightarrow V_A=1.60\,\mathrm V,\quad V_B=6.60\,\mathrm V.

(b) Interpretation and further calculation

IAB=VB/3=2.20A=3−VA/2.I_{AB}=V_B/3=2.20\,\mathrm A=3-V_A/2.P5=5IAB=11.0Wdelivered.P_{5}=5I_{AB}=11.0\,\mathrm W\quad\text{delivered}.

Physical check

The unknown current through an ideal voltage source is found from a node balance after the potentials are solved.