EE204
Lecture 02
Kirchhoff’s Current and Voltage Laws
Kirchoff’s Current Law (KCL).
The sum of currents entering a node is equal to the sum of currents leaving that node.
i1 + i4 = i2 + i3 + i5
Equivalent statement of KCL:
The algebraic sum of currents entering a node is equal to zero.
i1 − i2 − i3 + i4 − i5 = 0
Figure 8
Example 4:
Calculate the unknown currents in the following circuits.
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Figure 9
Solution:
a) KCL at node (a)
⇒
i1 = 2 + 4 = 6 A
b) KCL at node (a)
⇒
3 + i1 = 1
⇒
i1 = −2 A
⇒
3 + i1 − 1 = 0
⇒
i1 = −2 A
⇒
i1 − 4 + 2 = 0
⇒
i1 = 2 A
0.5 − i2 − 2 = 0
⇒
i2 = −1.5 A
Alternatively
KCL at node (a)
c) KCL at node (b)
KCL at node (c)
⇒
Check KCL at node (a) ⇒
−i1 + 4 + i2 − 0.5 = −(2) + 4 + (−1.5) − 0.5 = −4 + 4 = 0
KCL is also applicable to a closed area (super node).
The algebraic sum of currents entering a super node is equal to zero.
i1 + i2 − i3 + i4 − i5 = 0
Figure 10
Example:
Calculate the currents i1 and i2 in the circuit shown below:
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Figure 11
Solution:
KCL at super node 1
⇒
3 − i1 = 0
⇒
i1 = 3 A
KCL at super node 2
⇒
i2 = 0
⇒
i2 = 0 A
Ω
Ω
Ω
Ω
Figure 12
Kirchoff’s Voltage Law (KVL):
The algebraic sum of voltages around any closed circuit is equal to zero.
CW = clockwise
&
−v1 − v2 + v3 − v4 + v5 = 0 (1)
+v1 + v2 − v3 + v4 − v5 = 0 (2) [same as
⇒
⇒
KVL around circuit 1 (CW)
KVL around circuit 1 (CCW)
(1)]
CCW = counterclockwise
⇒
KVL around the outer circuit (CW)
KVL around circuit 2 (CW)
⇒
−v6 + v8 + v3 − v4 + v5 = 0
−v6 + v7 = 0
⇒
(3)
v6 = v7 (parallel elements)
Figure 1
Alternative KVL Statement:
The algebraic sum of voltages between two nodes is independent of the path taken
from the first node to the second node.
path1&2
KVL
Node a
→
Node b
⇒
Node b
⇒
+v2 + v1 = + v3 − v4 + v5
(4)
[same as
(5)
[same as
(1)]
path 2&3
KVL
(3)]
Node a
→
+v3 − v4 + v5 = −v8 + v6
+ V8
-
Path
2
+
V4
+
-
V5
Pa
-
-
V2
V1
V7
-
+
V6
+ V3
a
Path 1
c
th
3
b
Figure 2
Example:
Calculate the unknown voltages in the given circuit.
Ω
Ω
Ω
Figure 3
Solution:
Applying KVL:
Right-hand circuit (CW)
⇒
−(7) + v1 + (−1) + 10 = 0
⇒
v1 = −2V
Right-hand circuit (CCW)
⇒
+(7) − (10) − (−1) − v1 = 0
⇒
v1 = −2V
⇒
+v1 = +(7) − (10) − (−1)
⇒
v1 = −2V
path1&2
Node a
→
Node b
Same answer in all cases.
Left-hand circuit (CW)
path 3&4
Node a
→
Node c
⇒
+(7) − (v2 ) = 0
⇒
⇒
+v2 = +7
⇒
v2 = 7V
v2 = 7V
Same answer in both cases.
Ω
Ω
Ω
Ω
Ω
Ω
Figure 4
Example: (KVL)
Determine voltages vx, vy, vz in the circuit of fig….by applying KVL.
Solution:
KVL around the loop abcfa
−vz + (−6) + 3 − 2 = 0
⇒ vz = −6 − 2 + 3 = −5V
(1)
KVL around the loop fedef
2 + v y + (−1) = 0
⇒ v y = −2 + 1 = −1V
(2)
KVL around the loop bcdeb
−3 + 2 + vx − v y = 0
(3)
To get vx we can substitute vy from (2) into (3) to get:
vx = +3 − 2 + v y = 1 + (−1) = 0
⇒ vx = 0V
Note: We can also apply KVL around the loop febcdf to get vx directly:
2 − 3 + 2 + vx + (−1) = 0
⇒ vx = −2 + 3 − 2 + 1 = 0V