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 © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

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. McGraw-Hill © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

7-1: Series Voltage Dividers The Largest Series R Has the Most V. V1 = R1 RT × VT = 1 kW 1000 kW × 1000 V = 1 V V2 = 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; V2 is almost equal to the whole VT. KVL check: 1 V + 999 V = 1000 V Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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.

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. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

7-3: Current Division by Parallel Conductances For any number of parallel branches, IT is divided into currents that are proportional to the conductance of the branches. For a branch having conductance G: I = × IT G GT

7-3: Current Division by Parallel Conductances G1 = 1/R1 = 1/10 Ω = 0.1 S G2 = 1/R2 = 1/2 Ω = 0.5 S G3 = 1/R3 = 1/5 Ω = 0.2 S Fig. 7-5: Current divider with branch conductances G1, G2, and G3, each equal to 1/R. Note that S is the siemens unit for conductance. With conductance values, each branch I is directly proportional to the branch G. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

7-3: Current Division by Parallel Conductances The Siemens (S) unit is the reciprocal of the ohm (Ω) GT = G1 + G2 + G3 = 0.1 + 0.5 + 0.2 GT = 0.8 S I1 = 0.1/0.8 x 40 mA = 5 mA I2 = 0.5/0.8 x 40 mA = 25 mA I3 = 0.2/0.8 x 40 mA = 10 mA KCL check: 5 mA + 25 mA + 10 mA = 40 mA = IT

7-4: Series Voltage Divider with Parallel Load Current Voltage dividers are often used to tap off part of the applied voltage for a load that needs less than the total voltage. Fig. 7-6: Effect of a parallel load in part of a series voltage divider. (a) R1 and R2 in series without any branch current. (b) Reduced voltage across R2 and its parallel load RL. (c) Equivalent circuit of the loaded voltage divider. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

7-4: Series Voltage Divider with Parallel Load Current V1 = 40/60 x 60 V = 40 V V2 = 20/60 x 60 V = 20 V V1 + V2 = VT = 60 V (Applied Voltage) Fig 7-6

7-4: Series Voltage Divider with Parallel Load Current The current that passes through all the resistances in the voltage divider is called the bleeder current, IB. Resistance RL has just its load current IL. Resistance R2 has only the bleeder current IB. Resistance R1 has both IL and IB. Fig. 7-6

7-5: Design of a Loaded Voltage Divider Fig. 7-7: Voltage divider for different voltages and currents from the source VT. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

7-5: Design of a Loaded Voltage Divider I1 through R1 equals 30 mA I2 through R2 is 36 + 30 = 66 mA I3 through R3 is 54 + 36 + 30 = 120 mA V1 is 18 V to ground V2 is 40 − 18 = 22 V V3 is 100 V (Point D) − 40 = 60 V

7-5: Design of a Loaded Voltage Divider R1 = V1/I1 = 18 V/30 mA = 0.6 kΩ = 600 Ω R2 = V2/I2 = 22 V/66 mA = 0.333 kΩ = 333 Ω R3 = V3/I3 = 60 V/120 mA = 0.5 kΩ = 500 Ω NOTE: When these values are used for R1, R2, and R3 and connected in a voltage divider across a source of 100 V, each load will have the specified voltage at its rated current.