This document contains tutorials and practice problems related to applying Kirchhoff's Current Law (KCL) and Kirchhoff's Voltage Law (KVL) to solve for unknown voltages and currents in electric circuits. It includes 6 sample circuit problems where the key steps of writing KCL and KVL equations at nodes and around loops are demonstrated to find values such as the voltage across a resistor (vx), total circuit current (ix), or power dissipated by individual resistors. Diagrams illustrate the node-branch analysis method and solutions are provided for determining variables like resistance (RA) from given voltages and currents.

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Tutorial 3: Focus
 KCL and KVL Problems

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T3_Problem_1
 Find vx and ix :
Note: This is Practice problem 3.2 on page 43, Figure 3.7 in the Ref 1 book (by Hayt)
-4 V and -400 mA
3 + 1 + vx = 0
vx = -4 V
ix = -400 mA
Start from the node A and move in the clockwise direction till reaching A again
ix = vx / 10 = -4/10
A

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T3_Problem_2
 How many nodes and braches are there?
 Find RA =
Note: This is Practice 3.1 on page 42, Figure 3.4 in the Ref 1 book (by Hayt)
1 Ω
8 + 13 = iRA + ix
Total current entering the node A = Total current leaving the node A
= 3A
8 A
Nodes = 3
Branches = 5
18 V
Node A
At Node A:
iRA
21 = iRA + 3
iRA = 18
RA = VRA / iRA = 18/18
RA = 1 ΩGND or Reference Node

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T3_Problem_3
 Find vae & vec : 14 V & -10 V
vae + 10 - 24 = 0
Use of double subscript notation
vae = 14 V
vec + 4 + 6 = 0
vec = - 10 V

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T3_Problem_4
 Find vx : 8 V
Note: This is
Example problem
is 3.4 on page 45,
Figure 3.10 in the
Ref 1 book (by
Hayt)
-60 + v8 + v10 = 0
v8 = 5 * 8 = 40 V
-60 + 40 + v10 = 0
v10 = 20 V
-v10 + v4 + vx = 0
-20 + 12 + vx = 0
vx = 8 V
i10 = v10 /10
i10 = 20 /10
i10 = 2 A
i10 + i4 = 5 A
i4 = 5 – 2 = 3 A
v4 = 3 * 4 = 12 V

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T3_Problem_5
 Find vx : 12.8 V
Note: This is
Practice problem is
3.4 on page 46,
Figure 3.11 in the
Ref 1 book (by
Hayt)
-30 + v8 + v10 = 0
v8 = 2 * 8 = 16 V
-30 + 16 + v10 = 0
v10 = 14 V
-v10 + v2 + vx = 0
-14 + 1.2 + vx = 0
vx = 12.8 V
i10 = v10 /10
i10 = 14/10
i10 = 1.4 A
i10 + i2 = 2 A
i2 = 2 – 1.4 = 0.6 A
v2 = 0.6 * 2 = 1.2 V
v10
v8
vx
i2
i10
v2+
+
+ +
-
-
--

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T3_Problem_6
 Power absorbed: P30 = 0.768 W, P12V = 1.92 W, P8 = 0.2048 W, P7 =0.1792 W
P4vx = -3.072 W
Note: This is Practice problem 3.6
on page 48, Figure 3.14 in the Ref 1
book (by Hayt)
-vx - 12 + v8 + v7 + 4vx = 0 vx = -30 * i = -30 i
v8 + v7 + 3vx = 12
i
v8 = 8 * i = 8 i
v7 = 7 * i = 7 i
8i+ 7i-90i= 12
-75i= 12, i = -0.16 A
P30 = i2 * 30 = (0.16)2 * 30 = 0.768 W
P12V = v* i = (12 * 0.16)= 1.92 W
P8 = i2 * 8 = (0.16)2 * 8= 0.2048 W
P7 = i2 * 7 = (0.16)2 * 7= 0.1792 W
P4vx = 4 vx * i = -30*4*0.16* 0.16
= -3.072 W

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References
Ref 1 Ref 2