Voltage/current dividers
Trivial to understand, but still very useful.
Use in combination with equivalent resistances to quickly find a
particular voltage or current in a circuit.
Voltage divider
VS +
–
R1
iS
R2
R3
+
vR2
–
Want to know vR2.
Easily solved with KCL, KVL, &
equivalent resistances.
=
=
=
=
+
+
+
+
That’s it.
The total voltage across a series string is divided
among the resistors according to a simple ratio.
EE 201
voltage/current dividers – 1
vR1
+
–
R1
VS +
–
R2
R3
–
+
vR2
Easy to find the other voltages, too.
=
–
vR3
+
=
+
+
+
+
Put in some values.
+
+
15 V –
5 kΩ
–
15 kΩ
10 kΩ
–
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vR1
vR3
=
+
vR2
=
+
+
+
+
(
)= .
(
)= .
(
)=
–
+
=
+
+
voltage/current dividers – 2
Current divider
Same idea, but with current.
IS
R1
R2
iR2
R3
+
vR
–
=
=
=
=
+
+
Want to know iR2.
Easily solved with KCL, KVL, &
equivalent resistances.
+
+
The total is divided according a simple ratio determined by
the resistors.
EE 201
voltage/current dividers – 3
The other currents are just as easy.
IS
R1
iR1 R2
iR2 R3
=
iR3
=
+
+
+
+
Plug in some numbers.
=
9 mA
3 k!
5 k!
iR1
iR2
15 k!
iR3
=
=
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+
+
+
+
+
+
(
)=
(
)=
(
)=
voltage/current dividers – 4
Example 1
24 !
R1
R2
VS +
–
12 V
R3
48 !
R4
36 !
+
In the circuit, find vR4.
vR4
36 !
–
R5
12 !
vR4 = vR3
Combine parallel combinations.
R12 16 !
VS +
–
12 V
R5
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12 !
Use voltage divider.
=
18 !
+
R34
vR4
–
=
+
+
+
+
(
)
= .
voltage/current dividers – 5
Example 2
IS
Find the equivalent
resistance in each branch.
IS
0.5 A
0.5 A
R2 120 !
R3 40 !
R5
iR3
60 !
R1 10 !
R13 40 !
R4 20 !
R45
R6 80 !
iR13
80 !
R6 80 !
In the circuit, find iR3.
Use a current divider to find iR13.
=
=
EE 201
+
+
=
+
+
Referring back to the original circuit,
iR13 divides between R2 and R3.
( .
)= .
=
+
+
( .
)= .
voltage/current dividers – 6
Example 3
R1 1 k!
VS
R2
+
–
3 k!
Without RL,
=
RL
In the simple divider circuit at
right, if RL is attached in parallel
with R2, the voltage across R1
doubles. What is the value of
RL?
From the two expressions for v’R1
+
+
With RL,
=
+
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+
+
=
=
=
Then
=
=
=
=
+
= .
voltage/current dividers – 7
Example 4
R1
IS
1 mA
iR1
R2
iR2
R3
iR3
From the current divider equation, we
know that the currents are proportional to
the inverse of the resistance in each branch.
:
:
=
:
For the simple current divider at right,
design it (i.e. choose resistor values)
so that the currents are in the ratio
iR1:iR2 :iR3 = 1:2:4 and the total power
dissipated in the resistors is 10 mW.
The equivalent resistance of
the three in parallel is
=
+
+
:
+
=
+
=
=
=
.
Therefore,
:
:
=
:
:
R2 = 2R3 and R1 = 2R2 (R1 = 4R3).
=
.
Find Req,
=
Finally:
R3 = 1.75Req = 17.5 kΩ; R2 = 2R3 = 35 kΩ; R1 = 2R2 = 70 kΩ
EE 201
voltage/current dividers – 8