Name: Francisco, Marion Angelo V. Date: Feb. 3, 2015Year/Section: Third Year Professor: Engr. GarciaSchedule:10am - 1pm Tuesday Rating:

Experiment No. 1

THEVENIN’S THEOREM

I. OBJECTIVES

1. To study the application of Thevenin’s Theorem in circuit problems.2. To be familiar with the use of Thevenin’s Equivalent Circuit in solving problems.

II. INSTRUMENTS AND MATERIALS

QTY UNIT ITEM SPECSOne(1) Pc. DC Power Supply 12VdcOne(1) Pc. Multi-tester(VOM)One(1) Pc. DC Ammeter 0-1AOne(1) Set Connecting wiresOne(1) Pc. Extension cord(optional)One(1) Pc. IC 7812One(1) Set Resistor 500ohm to 10k

III. PROCEDURES

1. Connect the circuit shown in Fig. 10-1. Refer to Appendix for DC supply set up.2. Measure the load current I L and record the result to Table 10-1.

3. Remove the RL from the circuit. Referring to Fig. 10-2, measure V TH . Record the result on Table 10-1.

4. Disconnect the DC supply and replace it by a short circuit. Referring to Fig. 10-4, by using an ohmmeter, measure the resistance RTH . Record RTH on Table 10-1.

5. Referring to Fig 10-1, compute for the Thevenin’s equivalent voltage, resistance and load current. Record the results as compared value in Table 10-1 and calculate the percent difference.

IV. CIRCUIT DIAGRAMS

Fig. 10-1

Fig. 10-2

Fig. 10-3

Fig. 10-4

Fig. 10-5 Thevenin equivalent circuit with the load connected.

V. DATA AND RESULTS

V TH RTH I L

Experimental Value

Computed Value

% Difference

VI. COMPUTATION

VII. OBSERVATION

From the experiment i observed that to obtain the voltage in a Thevenin Circuit you must first remove the load resistor (R1 ) then measure the open circuit voltage. The obtained voltage is also called Thevenin Voltage.

To determine the current flowing in the circuit, i observed that you must replace the load resistor with a short circuit then measure the current. You measure the values for Thevenin voltage and short circuit current to determine the Thevenin Resistance or (RTH) and using the formula RTH = VTH/ISC. You can also determine the Thevenin resistance by replacing the voltage source with a short circuit then measure the resistance

You can apply Thevenins Circuit if there is a variable part of the circuit and the other elements are fixed or constant.

VIII. CONCLUSION

We therefore conclude that using a Thevenins theorem you can simply almost every complex circuit into a circuit that has only a load resistor and a single voltage source. By using Thevenin theorem it is now easier for you to distinguish the current and voltage flowing in the circuit. It is also more convenient and easier to achive different values for the circuit just by replacing your load resistor with different values.

Thevenin theorem is also only useful for "linear" type circuit and having ohmic components such as resistors, if you apply non ohmic components such as inductors and capacitors it will be harder for you to simplify and compute for the short circuit current and Thevenin voltage of the Circuit.

You can determine the Thevenin equivalent of the circuit by defining the problem and analyzing the circuit. If there are no independent sources you must apply your own voltage/current source to the circuit and then determine the short circuit current and Thevenin Voltage. Then you can now get the Thevenin Resistance then simplify the circuit.

IX. QUESTIONS / PROBLEMS

1. Under what conditions may Thevenin’s theorem be used to advantage?

= Thevenin's theorem is very useful when applied to linear circuits. These are the circuits which has resistors and bilateral networks where current to an element is not suspectible to change. It is also useful to circuits that has ohmic components such as resistors and it will fail if you add non ohmic components such capacitors , inductors etc. because the process will be more complicated and harder to find the required values.

2. What is meant by the open –circuit voltage? The resistance network “looking back into the network”. In the Thevenin’s theorem?

Open - circuit voltage = or OCV or Voc is the voltage which is not connected to any load in a circuit. It represents the voltage's full value due to the absence of a load it will share voltage with. It also operates on zero current.

The looking back resistance of a network is the equivalent electrical resistance of the network when someone looks back into the network from the terminals where said branch is connected.

3. Two batteries A and B are connected in parallel to supply a current of 160 amperes to load a resistor RL. The open-circuit terminal voltage of a battery A is 123 volts and its internal resistance is 0.12 ohm. If the internal resistance of a battery B is 0.15 ohm and it delivers 90 amperes, calculate (a) the open-circuit terminal EMF of a batter B; (b) the total power taken by the load resistor RL.

Given: Total Current = 160ABattery B Current= 90AVth of Battery A = 123VInternal Resistance of A = 0.12ΩInternal Resistance of B = 0.15 Ω

Find: V and P

Solution: 100-90A-70A= Battery A Current

0.15(90)-13.5V= Voltage rise at B

117 + 13.5 = 131.5 V

EMF of Battery B = 131.5 V

100 A (117.6V) = 18.720kW

PRL = 18.720kW