Exp 3

Electric lab

: Objective . to learn building the circuit-1 to learn measure the current and the -2 voltage in the circuit which contain on more . than one loop to verify node voltage and mesh current -3 . method and super position principle . to learn use this method on the circuit-4 using the pspice programs in paint the-5 . circuit

: Theory node voltage :the sum current inside and -1 outside in the one node equal zero ,(we can .) measure the voltage about this method I1 ± I2 ± I3 =0 =0

, I=V/R , V1/R1 ± V2/R2 ±V3/R3

mesh current : the sum voltage in the loop -2 equal zero ,(we can measure the current about .) this method V1 ± V2 ±V3 =0

, V=I*R , I1*R1 ± I2*R2

±I3*R3 =0

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Electric lab

super position : must be the circuit -3 independent source put the short circuit consideration about voltage source And put the open circuit consideration current , source . And then sum the two answer

: ( Result and diagram (data of table :Table 1 values Resisto Nominal values Measured values )rs (Ω

V )dc(v

)I dc (A

)V dc (V )I dc (A

R1

2.96

0.436

3.06

0.59

R2

12.9 6

2.759

12.89

2.75

R3

7.02 9

3.195

7.21

2.28

;Table 2 1

Pspice value

V dc )(V

I dc )(A

Electric lab

Measured quantities Partial voltage acrossMeasured quantities Total voltage )resistor (v PartialCURRENT across resistor Total current Measured calculated pspice )across resistor (mA

Resist )or (Ω Resist Measured )or (Ω 20v 10v 20v

R1 R1 R2 R2 R3 R3

10v

8.02

5.06 1.42 0.914.9 2.13 7 3.2 21.3 0.45 5.06 0.97

2.31

calculated 10v

20v

20v

10v

8.19

5.03 1.2 0.914.9 2.11 7 3.18 2.11 0.45 5.03 1.04

2.28

pspice 10v

20v

20v

10v

9.2

)(v across resistor )(mA .Calc pspic Meas .Meas .Calc epspic . 2.96 2.85 3.975 e

5.225 1.20 0.52 14.8 -5 0.768 12.84 1.81 3.144 2.75 1.81 0.38 5.225 7.19 0.82

:Table 3

1

2.375

3.28

0.52 12.86

0.437 12.99

2.73 7.14

2.764 7.035

3.32

3.195

Electric lab

: Calculation )Use mesh method :( table 1 Loop 1; -10+6.8k I1 +4.7 I1 -4.7 I2 -20=0 )I1 -4.7 I2=10 …………….(1 11.5 I2 -4.7 I1 +2.2I2+20=0 4.7 )4.7I1 +6.9 I2=-20………………(2)I1 +16.88 I2 =-48.92 11.5-( 2.44* )2( )I2=-38.92 12.18(= )1(+)2( I2=3.195 , I1=0.436 IR1 =0.436 mA ;V R1 =0.436*6.8K=2.96 V IR2=3.195-.436=2.759 mA ;VR2=2.759*4.7K =12.96 V IR3=3.195 mA

;VR3=3.195*2.2K=7.029V

Table (2+3): using super position Req=

+6.8=8.3Ω

Using voltage divider VR1=8.19v ;I R1 =8.19/6.8=1.2 mA VR2=VR3=2.13v ;I R2 =2.13/4.7=.453 mA ;I R3=2.13/2.2=0.97

Req=

+4.7=6.28Ω

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Electric lab

Using voltage divider VR2=14.96v ;I R1 =14.96/4.7=3.18 mA VR1=VR3=5.03v ;I R1=5.03/6.80.739 mA ;I R3=5.03/2.2=2.28 And then sum the voltage and current in the two condition VR1=2.85v ,IR1=0.52mA VR2=12.84v,IR2=2.73mA VR3=7.19v ,IR3=3.23mA : Experiment circuit

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Electric lab

:Conclusion exist the more than one method until-1 . calculation the voltage and current super position use only if the circuit -2 contain on dependant source (current or .) voltage

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