1. Voltage Dividers and Current Dividers
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

2. 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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3. 7-1: Series Voltage Dividers
The Largest Series R Has the Most V.
VR1 = R1 RT
× VT =
1 kW
1000 kW
× 1000 V = 1 V
VR2 = R2 RT
× VT
999 kW
1000 kW =
× 1000 V = 999 V
Fig. 7-2a: Example of a very small R1 in series with a large R2; VR2 is almost equal to the whole VT.
KVL check: 1 V V = 1000 V
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4. 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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5. 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
VBG =
2.5 kW
20 kW
× 24 V = 3 V
VAG = 12 V
VCG =
1 kW
20 kW
× 24 V = 1.2 V
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6. 7-1: Series Voltage Dividers

7. 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:
Note that this formula can only be used for two branch resistances.
The largest current flows in the branch that has the smallest R.

8. 7-2: Current Dividers with Two Parallel Resistances
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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