Problem 1-6: Single Node-Pair Analysis


Find  I_1 using single node-pair analysis (do not reduce the circuit).
Problem 1-6 - Single node-pair analysis
 Is_{1}=1A,\, Is_{2}=2A,\, Is_{3}=3A,\, R_1=2 \Omega,\, R_2=4\Omega and  R_3=4\Omega .


Solution
a) Redraw the circuit if necessary. Mark the voltage across nodes:

Single node-pair analysis - Redrawing the circuit
b) Apply KCL in one of nodes: \frac{V}{R_3}-Is_1 +Is_2+\frac{V}{R_2}-Is_3+\frac{V}{R_1}=0 \;(1) .

Please note that:

i) Voltage across each element is  V with the shown polarity.

ii) We do not use  I_1 , because it is unknown. We try to find any current in term of the main quantity, i.e.  V , which is unknown. Recall that main quantities are

  • node voltages in nodal analysis,
  • node voltage in single-node-pair analysis,
  • loop current in single loop analysis, and
  • mesh currents in mesh analysis.


iii) Since we are not using  I_1 , we do not care about its direction.

Eq. (1) gives  V= 2v .

c) Determine required quantity,  I_1 using main quantity, i.e.  V :

 V=-R_1\,I_1 \rightarrow I_1=-1A

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