8/13/2019 Lecture4 Voltage and Current Dividers

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Lecture 4

Voltage and Current Dividers

ECE 205

Prof. Ali Keyhani

Voltage Division

Provides a simple way to find the voltage

across an element in a series circuit

without solving the circuit equations

321:loopthearoundKVL vvvvs ++=

The elements are in series so the same current passes through each resistor:

321

321321

:Therefore

)(

RRR

vi

RRRiiRiRiRv

s

s

++=

++=++=

Using Ohms law the voltage across each resistor is found:

sss v

RRR

RiRvv

RRR

RiRvv

RRR

RiRv

++==

++==

++==

321

333

321

222

321

111

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Example 2

A) What is the voltage Vo when D1 and D0 are closed?

B) What is the voltage value is D0 is open

and D1 is closed?

VRR

Rv

k

kkk

O 75.3154

115:devisionVoltage

1R

321||5.3R

:aresresistancedividertheclosedareswitchesbothWhen

21

2

2

1

==+

=

=

==

VRR

Rv

k

kkk

O 5.7152

115:devisionVoltage

3R

321||5.3R

21

2

2

1

==+

=

=

==

Current Division

Provides a simple way to find the current

through an element in a parallel circuit

without solving the circuit equations

The elements are in parallel so the same voltage appears across each conductance:

Using Ohms law the current through each resistor is found:

321:AnodeatKCL iiiis ++=

321

321321

:Therefore

)(

GGG

iv

GGGvvGvGvGi

s

s

++=

++=++=

sss iGGG

GvGii

GGG

GvGii

GGG

GvGi

++==

++==

++==

321

3

33

321

222

321

111

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Current Division

From the equations a pattern is derived forthe resistors connected in parallel:

The source current divides among the parallelresistors in proportion to their conductances

divided by the equivalent conductances in

parallel connection.

Total

EQ

k

k iG

Gi

=

ruledivisioncurrentfor theexpressiongeneralThe

Example 3

Find the current ix.

Solution: The circuit is divided into two

paths. Then the current division rule isapplied to the equivalent circuit.

sssx iRR

Ri

RR

Ri

GG

Gi

+=

+

=

+=

21

2

21

1

21

1

11

1

A25.1567.620

67.6 =+

=xi