1.3.3 Current Divider

Chapter 1.3.3 Current Divider

Physics Lecture Notes – Phys 395 Electronics Book
Pages 176
Views 2,363
Downloads : 6 times
PDF Size : 906.7 KiB

Summary of Contents

Physics Lecture Notes – Phys 395 Electronics Book

  • CHAPTER 1. DIRECT CURRENT CIRCUITS11Vout =R2R1 + R2Vin.(1.12)Example: Determine an expression for the voltageV2 on the voltage divider infigure 1.4.V2V+R1R3R2Figure 1.4: Example voltage divider.We take the bottom line in figure 1.4 to be at ground and define the currentflowing betweenV2 and ground to be I. Ohm’s law givesV2 = IR23,whereR23 =R2R3R2 + R3.(1.13)Applying Kirchoff ’s voltage law for the input source givesV = IR,whereR = R1 + R23.(1.14)Combining the above two results and solving forV2 leads toV2 =R23RV =R2R3R2+R3R1 +R2R3R2+R3V(1.15)=R2R3R1R2 + R2R3 + R3R1V.(1.16)1.3.3Current DividerConsider the current divider shown in figure 1.3b. The source current is divided betweenthe two resistors and is given byIin =I1 +I2 =V/R1 +V/R2. The voltage at the outputisV =IoutR2. The output current from the current divider is thusIout =R1R1 + R2Iin.(1.17)Example: Determine an expression for the currentI3 through the resistorR3in the circuit shown in figure 1.5.The current I is divided amongst the three resistors and hence we use our expres-sion for resistors in parallel