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## Q. 1.12

Consider the following circuit.Find the value of$i_{x}$.

## Verified Solution

First, we label the nodes, define the directions of the currents, and define the voltages in accordance with the sign convention, as follows.

Note that we again define node 2 as the combination of two intersection points. Besides,
in order to simplify the solution, we do not define voltage variables separately as they
are already related to the currents via Ohm’s law. Using KVL, we derive
• KVL(1 → 2 → 1): $4i_{x} − 4i_{y}= 0 \longrightarrow i_{x} = i_{y}$,
• KVL(1 → 2 → 3 → 1): $4i_{x} − 12i_{z} − 20 = 0 \longrightarrow i_{z} = i_{x}∕3 − 5∕3$.
Then, using KCL, we obtain
• KCL(2): $i_{x} + i_{y} + i_{z} + 4 = 0$.
Finally, we have
$i_{x}+ i_{x} + i_{x}∕3 − 5∕3 + 4 = 0$
$7i_{x}∕3 = −7∕3 \longrightarrow i_{x} = −1 A$.

In solving this example, we further note the following.
• The voltage across the current source is unknown.Hence, it is not useful to write KVL for 3 → 2 → 3.
• The current across the voltage source is unknown.Hence, it is not useful to write KCL at 1 or 3.
In general,we avoid writing KCL at a node, to which a voltage source is connected, unless
it is mandatory to find the current through the voltage source. In addition, applying KVL
in a mesh containing a current source is usually not useful, unless the voltage across the
current source must be found