05-Basic Laws Part 2
Text: Chapter 2.4
ECEGR 210
Electric Circuits I
Overview
• Kirchhoff’s Voltage Law
• Kirchhoff’s Current Law
2
Introduction
• Last lecture left off with trying to analyze circuits
with two resistors
• To solve, apply Kirchhoff’s Law(s)
V1
12V
+
-
V2
+
+
-
3W
5W
5A
Dr. Louie
I1
3W
I2
5W
3
Kirchhoff’s Laws
• Named after Gustav Robert Kirchhoff
• Two laws:
 Kirchhoff’s Voltage Law (KVL)
 Kirchhoff’s Current Law (KCL)
• Both can be derived from Maxwell’s
Equations
Dr. Louie
4
Kirchhoff’s Laws
• Start with KVL
• KVL: sum of all voltages around a loop (closed
path) is zero
M
v
m1
m
0
 M: number of voltages in the loop
+
-
A
+
-
B
Dr. Louie
5
KVL
• Sum of voltage around a loop must equal zero
• Pay careful attention to polarity!
• Which circuit(s) obey Kirchhoff’s Voltage Law?
+
-
-
10V 12V
+
+
-
-
Dr. Louie
12V
12V
+
-
12V
+
+
-
12V
6
KVL
• Sum of voltage around a loop must equal zero
• Pay careful attention to polarity!
+
-
-
12  10  0
10V 12V
+
+
-
-
12  12  0
Dr. Louie
12V
12V
+
-
12V
+
+
-
12V
12  12  0
7
KVL
• Example
 Loop 1: 0 = -VA + VB + VC or VA = VB + VC
 Loop 2: 0 = -VA + VB - VD or VA = VB - VD
 Loop 3: 0 = -VC - VD or VC = VD
+
+
-
B
A
+
+
-
C
+
+
D
+
Loop 1
+
-
B
A
+
B
A
-
-
+
C
-
+
D
Loop 2
C
-
+
D
Loop 3
Dr. Louie
8
KVL
• Identify which loop to check
• Start at a node in the loop
• Sum voltages in loop accounting for polarity
 If a “+” is encountered, add the voltage
 If a “-” is encountered, subtract the voltage
• Check sum
 If equal to 0, then KVL is obeyed
 If not equal to 0, then there is an error in the
circuit
• Repeat for additional loops until all elements
considered
Dr. Louie
9
KVL
• Does this circuit obey KVL?
-
16V
+
12V
+
-
4V
-
+
6V
+
-
+
+
-
-
Dr. Louie
3V
3V
10
KVL
• Consider loop on the left
-12 – 4 + 16 = 0 (KVL is obeyed)
-
16V
+
12V
+
-
4V
-
+
6V
+
-
+
+
-
-
3V
3V
x
Dr. Louie
11
KVL
• Now try right loop
-16 + 4 + 6 +3 + 3= 0 (KVL is obeyed)
-
16V
+
12V
+
-
4V
-
+
6V
+
-
+
+
-
-
3V
3V
x
Dr. Louie
12
KVL
• We could have considered
-12 + 6 + 3 + 3= 0 (KVL is obeyed)
-
16V
+
12V
+
-
4V
-
+
6V
+
-
+
+
-
-
3V
3V
x
Dr. Louie
13
KVL
• What must VR be?
VR
12V
+
2W
-
+
+
-
-
Dr. Louie
3V
14
KVL
• What must VR be?
-12 + VR + 3 = 0
VR = 9V
• We could also compute the current using Ohm’s
Law (but it has added steps)
9 = 2I
+
I = 4.5A
2W
VR
2x4.5 = 9V
12V
+
+
-
-
Dr. Louie
3V
15
Combining Voltages
• KVL suggests that voltages in series can be
simply combined
I
V1
+
-
V
V2
I
+
V1 + V2
+
-
+
-
+
V
-
-
Dr. Louie
16
Combining Voltages
• In fact, any series elements (passive or active)
can be arbitrarily rearranged
 Polarity must not be changed
I
+
V1
R
V
+
-
I
+
-
R
V2
I
+
V
V1 + V2
I
+
-
-
-
Dr. Louie
17
Example
• Find the current through the resistor
I
5V
+
-2V
+
-
6W
-
Dr. Louie
18
Example
• Find the current through the resistor
 Combine voltage sources: 5 – 2 = 3V
 Apply Ohm’s Law: I = 3/6 = 0.5A
I
I
5V
-2V
+
-
6W
3V
+
-
6W
+
-
Dr. Louie
19
Kirchhoff’s Laws
• Kirchhoff’s Current Law (KCL) is analogous to
Kirchhoff’s Voltage Law
• KCL: sum of all currents a node (closed
boundary) is zero
N
i
n1
n
0
node
 N: number of branches connected to the node
Dr. Louie
20
KCL
• Sum of currents into a node must equal zero
• Pay careful attention to polarity!
• Which circuit(s) obey KCL?
12A
10A
12A
12A 12A
Dr. Louie
-12A
21
KCL
• Sum of currents into a node must equal zero
• Pay careful attention to polarity!
• Which circuit(s) obey KCL?
12A
10A
12A
12A 12A
Dr. Louie
-12A
22
KCL
• Example
 Node 1: 0 = IA – IB or IA = IB
 Node 2: 0 = IB – IC + ID or IC = IB + ID
 Node 3: 0 = IC – IA – ID or IC = IA + ID
2
1
B
C
A
D
3
Dr. Louie
23
KCL
• Identify which node to check
• Sum currents into/out of node, accounting for polarity
 If the current enters the node, the current is
positive
 If the current leaves the node, the current is
negative
• Check sum
 If equal to 0, then KCL is obeyed
 If not equal to 0, then there is an error in the
circuit
• Repeat for additional loops until all elements
considered
Dr. Louie
24
KCL
• Does this circuit obey KCL?
16A
4A
12A
-16A
4A
Dr. Louie
25
KCL
• Does this circuit obey KCL?
• Yes




0
0
0
0
=
=
=
=
12 + 4 – 16
16 + –16
4–4
-12 – 4 - -16
16A
4A
12A
-16A
4A
Dr. Louie
26
KCL
• What must I1 be?
5A
I1
4Ω
Dr. Louie
2A
27
KCL
• What must I1 be?
I1 = 5 + 2 = 7A
5A
I1
4Ω
Dr. Louie
2A
28
Combining Current
• KCL suggests that current sources in parallel can
be simply combined
I
I1
I
+
I2 V
-
I1 + I2
Dr. Louie
+
V
-
29
Combining Current
• In fact, parallel elements can be arbitrarily
rearranged
 Polarity must not be changed
I
I1
R
V1
+
-
I
+
V1
-
V1
Dr. Louie
+
-
I1
R
+
V1
-
30
Example
• A node has three branches. The current into the
node from branch 1 is 1.5A; the current into the
node from branch 2 is 3A. What is the current
into the node from branch 3?
A: 1.5A
B: 4.5A
C: -1.5A
D: -4.5A
Dr. Louie
31
Example
• A node has three branches. The current into the
node from branch 1 is 1.5A; the current into the
node from branch 2 is 3A. What is the current
into the node from branch 3?
A: 1.5A
B: 4.5A
1.5A
1.5A
C: -1.5A
4.5A
=
D: -4.5A
-4.5A
3A
3A
Dr. Louie
32