Kirchhoff’s Voltage Law (KVL) KVL Problems Home Work Problems

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Session 5: Focus
 Kirchhoff’s Voltage Law (KVL)
 KVL Problems
 Home Work Problems

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Kirchhoff’s Voltage Law (KVL)

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Kirchhoff’s Voltage Law (KVL)
 Current is related to the charge flowing through a
circuit element
◦ Whereas voltage is a measure of potential energy difference
across the element
 There is a single unique value for any voltage in circuit
theory.
 Thus, the energy required to move a unit charge from
point A to point B in a circuit must have a value
independent of the path chosen to get from A to B
 KVL: The algebraic sum of the voltages around any
closed path is zero

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KVL
• If 1 C charge is carried from A to B through element 1,
the reference polarity signs for v1 show that v1 joules of
work is done
• If, instead, 1C charge is carried from A to B via node
C, then (v2 − v3) joules of energy will be spent
• So, v1 = v2 - v3

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KVL
v1 + v2 + v3 + … + vn = 0
 It follows that if we trace out a closed path, the
algebraic sum of the voltages across the individual
elements around it must be zero:

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Applying KVL to a Circuit
 One method that leads to least error while writing the KVL
equation is the following:
 Moving mentally around the closed path in a clockwise
direction and writing down directly the voltage of each element
whose (+) terminal is entered, and
 Writing down the negative of every voltage first met at the (−)
sign
Start from B and move
in the clockwise direction
-v1 + v2 – v3 = 0
Same as the previous result
v1 = v2 - v3

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KVL Problems

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Problem 1: KVL
 Find vx and ix :
Note: This is Example problem 3.2 on page 43, Figure 3.6 in the Ref 1 book (by Hayt)
12 V and 120 mA
- 5 – 7 + vx = 0
vx = 12 V
ix = 120 mA
Start from the node A and move in the clockwise direction till reaching A again
A
ix = vx / 100 = 12/100

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Problem 2: KVL
 Find VR3 :
Note: This is Example problem 2.9 on page 36, Figure 2.9 in the Ref 2 book (by Irwin)
20 V VR1 – 5 + VR2 – 15 + VR3 - 30 = 0
Start from the node a and move in the clockwise direction till reaching a again
18 V
12 V
VR1 + VR2 + VR3 = 50
18 + 12 + VR3 = 50
30 + VR3 = 50
VR3 = 20

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Problem 3: KVL
 For the single node-pair circuit find: iA iB iC :
Note: This is Practice problem 3.8 on page 51, Figure 3.18 in the Ref 1 book (by Hayt)
3 A, -5.4A, 6A
5.6 = iA + iB + iC + 2
Node A
Nodes: 2
Node pairs: 1
At Node A: (KCL)
3.6 = iA + iB + iC
vx= iA *18 = iC * 9
iA *18 = iC * 9
iC = 2 iA
iB = -0.1 vx
vx = - 10 iB
iB = - 1.8 iA
vx= iA *18 = - 10 iB
3.6 = iA – 1.8 iA + 2iA
3.6 = 1.2 iA
iA = 3 A
Node B

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Problem 4: KVL
 Find VR2 & vx : 32 V & 6 V
-VR2 + 12 + 14 + vx = 0
-32 + 12 + 14 + vx = 0
4 – 36 + VR2 = 0
vx = 6 V
Starting from node c
Within the left loop
VR2 = 32 V
Starting from node b
Within the right loop

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Home Work Problems

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S5_HW_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
Start from the node A and move in the clockwise direction till reaching A again
A

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S5_HW_Problem_2
 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)

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S5_HW_Problem_3
 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)
v10
v8
vx
i2
i10
v2+
+
+ +
-
-
--

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S5_HW_Problem_4
 Power absorbed by 4vx source: -3.072 W
Note: This is Practice problem 3.6
on page 48, Figure 3.14 in the Ref 1
book (by Hayt)
i

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Session 5: Summary
 Kirchhoff’s Voltage Law (KVL)
 KVL Problems
 Home Work Problems

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References
Ref 1 Ref 2