Procedure: 1. Connect the circuit shown in Fig. below, the value of R=1k Q. Increase the value of voltage and measure current in each step, record it in table below. Frorm ohm's Law (Ginen R= IKAr) Votnge cv) Current (Amper) ImA 2mA 4v SV 4mA SmA GmA 9ltheory) ImA 2mA 3mA 4mA 6m A = Ivotts ) ImA Similarly 2volh = 2MA : 2. Repeat step (1) with change the value of resistance R=330 N, measure current and record it in table below. voitage(volt) 3v Current (Amper) 3.03MA 6 t6mA 9•0FMA 12 12MA 15 15mp 18'18mA 4v Sv 6v I (the oy> BosmA6.06mA 109MA|12l2m A1S-ISMA 18:18MA

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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The same solution, please, but in a nice line. Don't forget my name
Name: Morteza Ali Qasim
The group: B1
My bad hand line saved me. I want the same solution
in a beautiful line, please, a beautiful and tidy line. I
know a strange request, but I will give you an
admiration, don't forget my name to you, I respect
you and appreciate you, docfor, don't forget my name
in a paper
Procedure:
1. Connect the circuit shown in Fig. below, the value of R=1k Q. Increase the value
of voltage and measure current in each step, record it in table below.
From ohm's Law
(Ginen R= IKr)
Voltage Cv)
Current (Amper) ImA 2mA
2v
4v
4mA
SmA
6mA
Gltheory)
ImA
2mA
3mA
4mA
6m A
= Iveltsa ImA
Similarly
1=V = 2Vols
I K
ニ2MA-
2. Repeat step (1) with change the value of resistance R=330 2, measure
current and record it in table below.
voitage(voit)
Current (Amper) 3-03MA G t6mA| 9.0mA 12 12MA 15 15MA 1818m
2V
4v
Sv
BosmA6.06MA1.09mA1212m A1S-ISMA 18:18 mA
を
Ivolts
-3.03MA
330
1= 2volts
6 mA
330 V
3. Repeat step (1) with change the value of resistance R=680 2, measure
current and record
in table below.
Vo tnge Cvoik)
Iv
3V
4V
Culent( Ampes):43m 2·14m 4.41MA5.88mA7-35MA882MA
1 (thedug)
1.43mA 2.94mA 4.41MA 5880A735MA 6 82mA
Ivolt
680N
%3D
=149m A
1- 2volt
E2.94 m A
68ON
Transcribed Image Text:Name: Morteza Ali Qasim The group: B1 My bad hand line saved me. I want the same solution in a beautiful line, please, a beautiful and tidy line. I know a strange request, but I will give you an admiration, don't forget my name to you, I respect you and appreciate you, docfor, don't forget my name in a paper Procedure: 1. Connect the circuit shown in Fig. below, the value of R=1k Q. Increase the value of voltage and measure current in each step, record it in table below. From ohm's Law (Ginen R= IKr) Voltage Cv) Current (Amper) ImA 2mA 2v 4v 4mA SmA 6mA Gltheory) ImA 2mA 3mA 4mA 6m A = Iveltsa ImA Similarly 1=V = 2Vols I K ニ2MA- 2. Repeat step (1) with change the value of resistance R=330 2, measure current and record it in table below. voitage(voit) Current (Amper) 3-03MA G t6mA| 9.0mA 12 12MA 15 15MA 1818m 2V 4v Sv BosmA6.06MA1.09mA1212m A1S-ISMA 18:18 mA を Ivolts -3.03MA 330 1= 2volts 6 mA 330 V 3. Repeat step (1) with change the value of resistance R=680 2, measure current and record in table below. Vo tnge Cvoik) Iv 3V 4V Culent( Ampes):43m 2·14m 4.41MA5.88mA7-35MA882MA 1 (thedug) 1.43mA 2.94mA 4.41MA 5880A735MA 6 82mA Ivolt 680N %3D =149m A 1- 2volt E2.94 m A 68ON
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