Lecture4 Voltage and Current Dividers
Transcript of 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