Sample problems
Topic: KCL and KVL
Statement of problem
The variable resistor R in the circuit below is adjusted until
a
v
equals 60 V.
240V
R
45
Ω
10
Ω
180
Ω
18
Ω
12
Ω
a
v
+
−
Find the value of R.
Solution
The current flowing through the
12
Ω
resistor is easily obtained from
a
v
1
60
5
12
12
a
v
V
iA
===
ΩΩ
In this circuit, there are six branches in total, which are shown in the figure below.
1
5
=
2
i
3
i
4
i
5
i
6
i
1
n
2
n
3
n
Therefore, five currents are unknown. Together with the value of R, there are six
unknowns in total, so six equations are needed. However, only three nodes (see figure
above) are ready for INDEPENDENT KCL equations. Obviously, the other three
equations are KVL equations in three loops: the upper one, the lower left one and the
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 Spring '05
 Unknown
 Electrical Engineering, Harshad number, INDEPENDENT KCL equations, figure below. i3, Sample problems Topic

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