BULACAN STATE UNIVERSITYCITY OF MALOLOS, BULACAN

COLLEGE OF ENGINEERINGEE 311L ACTIVITY NO.6CIRCUITS WITH TWO POWER SOURCES

SUBMITTED BY: GROUP NO. 2LEADER: VALILA, MARY GRACE CATHERINE I.CO-LEADER: ENRIQUEZ, EPHRAEM MANUELMEMBERS: CABRERA, CLARISSE JOYCASTILLO, IAN RENNIERCALALANG, PAMELADOMINGO, ESTEFANY REMUELSANCHEZ, CLARENZVILLANUEVA, LEVIE JOY

C/Y/S: BS ECE 3BDATE SUBMITTED: OCTOBER 8, 2015SUBMITTED TO: ENGR. DANIEL A. CRISOSTOMO(INSTRUCTOR)1st SEMESTER / S.Y. 2015 2016

EE 311LACTIVITY NO.6CIRCUITS WITH TWO POWER SOURCES

I. OBJECTIVES

TO INVESTIGATE THE CIRCUITS WITH TWO POWER SOURCES. TO OBTAIN EVIDENCE FOR THE MATHEMATICAL RELATIONSHIP THAT HOLD FOR CIRCUITS WITH TWO POWER SOURCES.

II. MATERIALS NEEDED

POWER SOURCE: 6V (EB) AND 12V (EA) DC CONNECTING WIRES RESISTOR1 RESISTOR2 RESISTOR3

III. CIRCUIT DIAGRAM:

I1 I2 I3

VA VB

IV. MATHEMATICAL COMPUTATIONS: (SHOW COMPLETE SOLUTION ANY METHOD)

R1 = 10kOhms R2 = 5.6 kOhmsR3 = 15 kOhms

V1= 16.83 VV2= 5.99 V

USING NODAL ANALYSISVD = 0VA VD = 16.83 V VC VD = 5.99 VVA= 16.83 V VC = 5.99 VAT NODE B42VA 42VB + 75VC 75VB = 28 VB42VA + 75VC = 145 VB(42)(16.83) + 75(5.99) = 145 VB1156.11 = 145 VBVB = 7.97 V

VB = I3R3

ANS: I3 = .53 mAI3 = .53 mAV2 = I2R2

ANS: I2 = .35 mAI2 = .35 mA

V1 = I1R1

ANS: I1 = .90 mAI1 = .90 mA

ANSWERSI1 = .90 mAI2 = .35 mAI3 = .53 mAV1 = 8.9 VV2 = 1.98 VV3 = 7.97 V

V. EXPERIMENTAL VALUE

VA= 16.83 VVB= 5.99 V

R1 = 10kOhms R2 = 5.6 kOhmsR3 = 15 kOhms

I1 = 0.93 mAI2 = 0.38 mAI3 = 0.55 mAV1 = 9.16 VV2 = 2.134 VV3 = 8.12 V

VI. OBSERVATION/CONCLUSION:MEASURED VALUES:VA= 16.83 VVB= 5.99 V

R1 = 10kOhms R2 = 5.6 kOhmsR3 = 15 kOhms

EXPERIMENTAL VALUECOMPUTED VALUE

I10.93 mA.90 mA

I20.38 mA.35 mA

I30.55 mA.53 mA

V19.16 V8.9 V

V22.134 V1.98 V

V38.12 V7.97 V

CONCLUSION:

PROBLEM NO. 1 1. SOLVE I1, I2, AND I3 IN THE CIRCUIT SHOWN BELOW:

I1 I2 I3

KIRCHHOFFS LAW

A.) METHODS OF ELIMINATIONB.) METHODS OF SUBSTITUTIONC.) METHODS OF DETERMINANTS

BY MAXWELL MESH ANALYSIS (ANY METHOD) BY NODAL NOTE ANALYSIS BY SUPERPOSITION THEOREMGIVEN:VA= 15 VVB= 13 V

R1 = 100 Ohms R2 = 20 OhmsR3 = 10 Ohms

REQUIRED:I1, I2, and I3

SOLUTION: BY KIRCHOFFS LAWS BY MESH ANALYSIS BY NODAL ANALYSIS

VD = 0VA VD = 15 V VC VD = 13 VVA= 15 V VC = 13 VAT NODE BVA VB + 5VC 5VB = 10 VBVA + 5VC 6VB = 10 VBVA + 5VC = 16 VB15 + 5(13) = 16VB80 = 16 VBVB = 5V

V3 = I3R3

ANS: I1 = .1AANS: I2 = .4AANS: I3 = .5AI3 = .5AV2 = I2R2I2 = .4A

V1 = I1R1I1 = .1A

BY SUPERPOSITION THEOREM

WE DIVIDED THE CIRCUIT INTO TWO PARTS CIRCUIT A AND CIRCUIT B

I1 I2 I3 I1 I2 I3

CIRCUIT A CIRCUIT B

SOLUTION FOR CIRCUIT A

LOOP 1 LOOP 2

AT LOOP 1 AT LOOP 2110 I1 10 I2 = 15-10 I1 + 30 I2 = 0BY CALCULATIONI1A = 0.1406 A I2A = 0.0468 AI3A = I1A - I2A = 0.1406 A 0.0468 A = 0.0938 AI3A = 0.0938 A

SOLUTION FOR CIRCUIT B

LOOP 1 LOOP 2

AT LOOP 1 AT LOOP 2110 I1 10 I2 = 0-10 I1 + 30 I2 = -13

BY CALCULATIONI1B = -0.04 A I2B = 0.44 AI3B= I1B - I2B = -0.04 A + 0.44 A = 0.4 AI3A = 4 ACOMBINING THE RESULTS FOR CIRCUIT A AND B WE GET:I1A = 0.1406 A I1B = -0.04 A I2A = 0.0468 A I2B = -0.44 AI3A = 0.0938 I3B = 0.4 A

I1 = I1A + I1B I2 = I2A + I2B I3 = I3A + I3BI1 = 0.1406 0.04 I2 = 0.44 0.04 I3 = 0.0938 + 0.4I1 = 0.1 A I2 = 0.4 A I3 = 0.5 A

ANS: I1 = 0.1 A I2 = 0.4 A I3 = 0.5 A

PROBLEM NO. 2

2. IN THE CIRCUIT GIVEN IN PROBLEM NO. 1: SOLVE FOR I3, USING

A. THEVENINS THEOREMB. NORTONS THEORE

GIVEN:VA= 15 VVB= 13 V

R1 = 100 Ohms R2 = 20 OhmsR3 = 10 Ohms

REQUIRED:I3 USING NORTONS AND THEVENINS THEOREMSOLUTION: A. THEVENINS THEOREMB. NORTONS THEOREM

NORTON EQUIVALENT CIRCUIT

LOOP 1 LOOP 2

AT LOOP 1 AT LOOP 2110 I1 = 15 20 I2 = 13I1 = .15 A I2 = .65 A

IN = I1 I2 IN = 0.15 0.65 IN =.5RN = 100 Ohms + 20 Ohms = 120 Ohms

THE EQUIVALENT CIRCUIT WILL BE:

USING THE CURRENT DIVIDER FORMULA WE GETI3 = = 0.49 A

ANS: I3 = .5A