(Experiment NO: 3)The Current Divider CircuitProcedure:1. Connect the circuit shown in Fig. (3).2. For R2=2200 measure I, Iı, I2 and record in table. 1.3. Replace R2 by another one of 120Q and measure I, Ii, I2 and record intable. 1.4. Replace R2 by zero value(R2=0) (shorted) and measure I, I1, I2 and record in table. 1A.Rs4700AVs=+15VR12200R2=22001200Figure(3)Amp.R2=2202R2=1202R2=0I1I2Measured current table 1

Solution: Case 1: If R2=220-ohm. Applying KCL at node V,…

(Experiment NO: 3) The Current Divider Circuit Procedure: 1. Connect the circuit shown in Fig. (3). 2. For R2=2200 measure I, I1, I2 and record in table. 1. 3. Replace R2 by another one of 1200 and measure I, I1, 12 and record in table. 1. 4. Replace R2 by zero value(R2=0) (shorted) and measure I, I1, I2 and record in table. 1 Rs 4700 Vs= 15V R1 2200 R2= 2200 1200 Figure(3) Amp. R2=2202 R2=1202 R2=0 I I1 I2 Measured current table 1

(Experiment NO: 3) The Current Divider Circuit Procedure: 1. Connect the circuit shown in Fig. (3). 2. For R2=2200 measure I, I1, I2 and record in table. 1. 3. Replace R2 by another one of 1200 and measure I, I1, 12 and record in table. 1. 4. Replace R2 by zero value(R2=0) (shorted) and measure I, I1, I2 and record in table. 1 Rs 4700 Vs= 15V R1 2200 R2= 2200 1200 Figure(3) Amp. R2=2202 R2=1202 R2=0 I I1 I2 Measured current table 1

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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