# Problem 1-6: Single Node-Pair Analysis

Find $I_1$ using single node-pair analysis (do not reduce the circuit).

$Is_{1}=1A,\, Is_{2}=2A,\, Is_{3}=3A,\, R_1=2 \Omega,\, R_2=4\Omega$ and $R_3=4\Omega$.

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

b) Apply KCL in one of nodes:$\frac{V}{R_3}-Is_1 +Is_2+\frac{V}{R_2}-Is_3+\frac{V}{R_1}=0 \;(1)$.

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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