Lesson 5 – Voltage and Current Dividers (Section 25)(CLO 25)
This arguably is one of the most useful lessons the students will learn.
We will use this extensively to
solve dc and ac problems and to design filters.
Start by deriving the relationship.
Use KVL and write a loop equation
0
3
2
1
S
=
+
+
+

V
V
V
V
Replacing each with Ohm’s law
)
(
3
2
1
3
2
1
S
R
R
R
I
IR
IR
IR
V
+
+
=
+
+
=
Solving for the series current
)
/(
3
2
1
S
R
R
R
V
I
+
+
=
We can then solve for the voltage across each resistor
)
/(
3
2
1
1
S
1
1
R
R
R
R
V
IR
V
+
+
=
=
)
/(
3
2
1
2
S
2
2
R
R
R
R
V
IR
V
+
+
=
=
)
/(
3
2
1
3
S
3
3
R
R
R
R
V
IR
V
+
+
=
=
Show that the sum of the three voltage drops add to the voltage supplied.
3
2
1
S
V
V
V
V
+
+
=
Point out that each voltage drop is proportional to the ratio of the individual resistor to the sum of all
the resistance in that branch.
Do some examples – using the same circuit let R
1
= 1 k
Ω
, R
2
= 5 k
Ω
and R
3
= 4 k
Ω
and V
S
= 100 V.
We compute the value of the voltage across each resistor and then add them up to show that it indeed
adds to 100 V.
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 Spring '08
 MAURICIODEOLIVERIA
 Voltage divider, Resistor, Electrical impedance, RX 100Ω RX

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