EAST WEST UNIVERSITY

Course Name : Electric Circuits - I

Course Code : EEE 101 LAB

Section No : 02

Group No : 06

Experiment No : 02

Name of the

Experiment : KIRCHHOFF’S VOLTAGE AND CURRENT LAWS

Date of allocation : 03/02/2013

Date of submission : 10/02/2013

Submitted To : Mariam B. Salim

Student’s ID : 2013-1-80-022

Student’s Name : Md. Solayman Khan

Post lab questions:1. Submit the experimental datasheet.2. Theoretically calculate the voltage V1, V2, V3, V4 and the

currents I1, I2, I3, I4, I5 and I6 using nodal analysis.Use the measured values of the resistors in this calculation.

3. Compare the calculated values with the measured values and comment on any observed discrepancy.

4. Due to uncertainties and errors associated with measurements,experimental data dusually do not exactly satisfy theoretical equations.In case of this experiment,how much discrepancy can be considered acceptable?Give justification for your answer.

Answer to the que no. 1 :

Data Sheet:

R1() R2() R3() R4()32.7 32.5 48.1 98.9

V1(v) V2(v) V3(v) V4(v)1.65 -0.65 3.4 1.03

I1(mA) I2(mA) I3(mA) I4(mA) I5(mA) I6(mA)49.3 19.3 68.9 9.9 59.3 9.4

Objective:The main object of this experiment is to verify

kirchhoff's voltage and current laws .

Figure 1 :

Answer to the que no 2:

At node 1: -I5+I4+I1=0 ................ (i)

At node 2: I2-I4-I6=0 ................ (ii)

At node 3: I1+I2-I3= ................(iii)

At Loop 1-3-0-1:

V1+V3-VS1=0

I1×32.7+I3×48.1-5=0 I1=5−48.1×I 3

32.7 ............. (iv)

At Loop 3-2-0-3:

V2+VS2-V3=0

-I2×32.5+4-I3×48.1=0 I2=4−48.1 I 3

32.5 ............ (v)

From (iii),(iv),(v) =>

5−48.1×I 332.7 +4−48.1×I 3

32.5 =I3

162.5-1563.25×I3+130.8-1572.87×I3=1062.75×I3 I3= 293.43

4198.87=0.06985A=69.85mA ........(vi)

So,V3=I3R3=0.06985×48.1=3.36V ........(vii)

Place ,I3=0.06985mA at (iv) =>

I1= 5−48.1×0.0698532.7 =0.05016A=50.16mA ........(viii)

So,V1=I1R1=0.05016×32.7=1.64V ........( xi)

Place ,I3=0.06985mA at (iv) =>

I2= 4−48.1×0.0698532.5 =0.0197A=19.7mA ........(x)

So,V2= -I2R1= -0.0197×32.5= -0.64V ........(xi)

Now, At Loop 1-2-0-1:

V4+VS2+-VS1=0

V4+4-5=0 V4=1 ........(xii)

So,

I4=V 4R4 = 1

98.9=0.01011A=10.11mA ........(xiii)

From (i) => -I5+I4+I1=0

=>I5=I4+I1

=> I5=0.01011+0.05016=0.6027A=60.27mA ........(xiv)

From (ii) => I2-I4-I6=0

=>I6=I2-I4

=> I6=0.0197-0.01011=0.00959A=9.59mA ........(xv)

From (vi) to (xv),the values of I1, I2, I3, I4, I5, I6, V1,V2,V3,V4 are Calculated.

Here,

I1=50.16mA ; V1=1.64v

I2= 19.70mA ; V2= -0.64v

I3= 69.85mA ; V3=3.36v

I4= 10.11mA ; V4=1.00v

I5= 60.27mA

I6= 9.59mA

Answer to the que no 3: Comparison Between Calculated and Measured Values of Voltages :

V1(v) V2(v) V3(v) V4(v)Calculated

Values1.64 -0.64 3.36 1.00

Measured Values

1.65 -0.65 3.40 1.03

Comparison Between Calculated and Measured Values of Currents :

I1(mA) I2(mA) I3(mA) I4(mA) I5(mA) I6(mA)Calculated Values

50.16 19.70 69.85 10.11 60.27 9.59

Measured Values

49.30 19.30 68.90 9.90 59.30 9.40

Comment: All the measured values of voltage of resistors was a little bit greater than the calculated values.On the other hand, All the measured values of flow of current was a little bit smaller than the calculated values.Without this,i didn’t see any discrepancy between my measurement and calculation.

I think, the voltage of VS2 Was a little bit higher than 4V, So, all voltages became little bit high.

As voltage rised,So logically it is sure that the current should also be high but here all the measured values of currents are lower than the calculated values of currents.

Answer to the que no:4

The percentage of discrepancy that i saw between my equation and calculation is given below:

percentage of discrepancy of voltage at V1: 1.65−1.641.65

×100 % =0.60%

percentage of discrepancy of voltage at V2: −.65−(−.64)−.65

×100 % =1.53%

percentage of discrepancy of voltage at V3: 3.40−3.363.40

×100 % =1.17%

percentage of discrepancy of voltage at V4: 1.03−1.001.03

×100 % =2.91%

percentage of discrepancy of Current at I1: 50.16−49.3050.16

×100 % =1.71%

percentage of discrepancy of Current at I2: 19.70−19.3019.70

×100 % =2.03%

percentage of discrepancy of Current at I3: 69.85−68.9069.85

×100 % =1.36%

percentage of discrepancy of Current at I4: 10.11−9.9010.11

×100% =2.07%

percentage of discrepancy of Current at I5: 60.27−59.3060.27

×100 % =1.60%

percentage of discrepancy of Current at I6: 9.59−9.409.59

×100% =1.98%

In this experiment, i got higher discrepancy at R4.

And it is =: 1.03−1.001.03

×100 % =2.91%

So,i think for this experiment,less than 3.0% discrepancy can be considered acceptable.That is why,because,in this experiment, i didn’t face any discrepancy more than 3% .

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