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Chapter 1

Q. 1.11

Consider the following circuit, where a voltage source and a current source are connected to three resistors. Find the value i_{x}, that is, the current across the 3 Ω resistor.


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.

Then, using KVL and Ohm’s law, we have v_{x} = 3i_{x}, v_{y} = 6i_{x}, and
• KVL(1 → 2 → 3 → 1): −24 + v_{x} + v_{x} = 0 \longrightarrow v_{x} + v_{y} = 24V

Therefore, we have
3i_{x} + 6i_{y} = 24 \longrightarrow i_{x} + 2i_{y} = 8 A.
Furthermore, using KCL (see below for some details), we derive
• KCL(2):i_{x} − i_{y} − i_{z} = −2A.
Using v_{y} = v_{z} and i_{y} = v_{y}∕6 = v_{z}∕6 = i_{z}, we obtain
i_{x} − 2i_{y} = −2 A.
Finally, solving two equations with two unknowns, we get
i_{x} = 3 A.
In the above, we note that node 2 (as well as node 3) is defined simultaneously at two
intersections, and KCL is written accordingly by considering all entering and leaving currents, as follows.

This is a common practice in circuit analysis in order to reduce the number of equations.
Specifically, intersections without a component between them can be considered as a
single node to avoid writing redundant equations with redundant unknowns. On the
other hand, this not mandatory. For example, one can consider each intersection as a node, as follows.

In this case, we need to define a current i_{w} between nodes 2 and 4.Writing KCL at the nodes, we now have
• KCL(2): i_{x} − i_{y}− i_{w} = 0,
• KCL(4): i_{w} + 2 − i_{z} = 0.
Obviously, when these equations are combined (directly added), we arrive at the same
equation in the original solution,
i_{x} − i_{y} − i_{z} = −2 A.