Lecture #3. OUTLINE KCL; KVL examples Dependent sources. Using Kirchhoff’s Voltage Law (KVL). Use reference polarities to determine whether a voltage is dropped – with no concern about actual voltage polarities. Consider a branch which forms part of a loop:. – v 2 +. + v 1 _.
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Consider a branch which forms part of a loop:
–
v2
+
+
v1
_
voltage
“drop”
voltage
“rise”
(negative drop)
loop
loop
(Conservation of energy)
Formulation 1:
Sum of voltage drops around loop
= sum of voltage rises around loop
Formulation 2:
Algebraic sum of voltage drops around loop = 0
Formulation 3:
Algebraic sum of voltage rises around loop = 0
+
va
_
+
vb
_
Applying KVL in the clockwise direction,
starting at the top:
vb – va = 0 vb = va
v2
v3
+
+
2
1
+
+
+
va
vb
vc

3
KVL ExampleThree closed paths:
b
a
c
Path 1:
Path 2:
Path 3:
Find v using KVL and KCL
Simplify a circuit before applying KCL and/or KVL:
Find I
I
R1
R1 = R2 = 3 kW
R3 = 6 kW
R3
R2
7 V
R4
R4 = R5 = 5 kW
R6 = 10 kW
R6
R5
Find Vx, Vy and Vz
A linear dependent source is a voltage or current source that depends linearly on some other circuit current or voltage.
We can have voltage or current sources depending on voltages or currents elsewhere in the circuit.
Here, the voltage V provided by the dependent source (right) is proportional to the voltage drop over Element X. The dependent source does not need to be attached to the Element X in any way.
The 4 Basic Dependent Sources is proportional to the voltage drop over Element X. The dependent source does not need to be attached to the Element X in any way.
Example is proportional to the voltage drop over Element X. The dependent source does not need to be attached to the Element X in any way. 2
Find is proportional to the voltage drop over Element X. The dependent source does not need to be attached to the Element X in any way. i2, i1 and io
i is proportional to the voltage drop over Element X. The dependent source does not need to be attached to the Element X in any way. D
20 W
10 W
–
+
200 W
2.4 A
80 V
–
+
5iD
Exercise