Chapter
7
Voltage Dividers
and Current Dividers
Topics Covered in Chapter 7
7-1: Series Voltage Dividers
7-2: Current Dividers with Two Parallel Resistances
7-3: Current Division by Parallel Conductances
7-4: Series Voltage Divider with Parallel Load Current
7-5: Design of a Loaded Voltage Divider
7-1: Series Voltage Dividers
 VT is divided into IR voltage drops that are proportional
to the series resistance values.
 Each resistance provides an IR voltage drop equal to its
proportional part of the applied voltage:
VR = (R/RT) × VT
 This formula can be used for any number of series
resistances because of the direct proportion between
each voltage drop V and its resistance R.
 The largest series R has the largest IR voltage drop.
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7-1: Series Voltage Dividers
 The Largest Series R Has the Most V.
VR1 =
R1
RT
× VT
1 kW
=
× 1000 V = 1 V
1000 kW
R2
× VT
RT
999 kW
=
× 1000 V = 999 V
1000 kW
VR2 =
KVL check: 1 V + 999 V = 1000 V
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Fig. 7-2a: Example of a very small R1 in
series with a large R2; VR2 is almost
equal to the whole VT.
7-1: Series Voltage Dividers
 Voltage Taps in a Series
Voltage Divider
 Different voltages are
available at voltage taps
A, B, and C.
 The voltage at each tap
point is measured with
respect to ground.
 Ground is the reference
point.
Fig. 7-2b: Series voltage divider with voltage taps.
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7-1: Series Voltage Dividers
Voltage Taps in a Series Voltage Divider
 Note: VAG is the sum of the
voltage across R2, R3, and R4.
 VAG is one-half of the applied
voltage VT, because
R2+R3+ R4 = 50% of RT
2.5 kW
× 24 V = 3 V
VBG =
VAG = 12 V
20 kW
1 kW
VCG =
× 24 V = 1.2 V
20 kW
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7-1: Series Voltage Dividers
7-2: Current Dividers with
Two Parallel Resistances
 IT is divided into individual branch currents.
 Each branch current is inversely proportional to the
branch resistance value.
 For two resistors,
R1 and R2, in parallel:
I1 
R2
 IT
R1  R 2
I1 
 Note that this formula can only be used for two branch
resistances.
 The largest current flows in the branch that has the
smallest R.
7-2: Current Dividers with
Two Parallel Resistances
 Current Divider
I1 = 4 Ω/(2 Ω + 4 Ω) × 30A = 20A
I2= 2 Ω /(2 Ω + 4 Ω) × 30A = 10A
Fig. 7-3: Current divider with two branch
resistances. Each branch I is inversely
proportional to its R. The smaller R has
more I.
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