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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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lecture 3
Lecture #3

OUTLINE

  • KCL; KVL examples
  • Dependent sources
using kirchhoff s voltage law kvl
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:

v2

+

+

v1

_

voltage

“drop”

voltage

“rise”

(negative drop)

loop

loop

formulations of kirchhoff s voltage law
Formulations of Kirchhoff’s Voltage Law

(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

  • Voltage rises are included with a minus sign.

Formulation 3:

Algebraic sum of voltage rises around loop = 0

  • Voltage drops are included with a minus sign.
a major implication of kvl
A Major Implication of KVL
  • KVL tells us that any set of elements which are connected at both ends carry the same voltage.
  • We say these elements are connected in parallel.

+

va

_

+

vb

_

Applying KVL in the clockwise direction,

starting at the top:

vb – va = 0  vb = va

kvl example

v2

v3

+ 

+ 

2

1

+

+

+

va

vb

vc

-

3

KVL Example

Three closed paths:

b

a

c

Path 1:

Path 2:

Path 3:

slide9

+

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

slide11

Dependent Sources

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.

slide12

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.

exercise

iD

20 W

10 W

+

200 W

2.4 A

80 V

+

5iD

Exercise
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