PART A – THEVENIN THEOREMS
Find the Thevenin and and Norton equivalent circuit to the left of terminals A-B for the network shown below. Connect the Norton equivalent circuit to the load and find the current in the 15Ω resistor. Show all the steps for the equivalent circuit.

Thevenin Theorem | Norton Theorem | VTH = 25Ω ×12 10Ω+25ΩVTH = 8.57 VRTH = [10Ω//25Ω] + 25Ω = [10 Ω×25Ω] + 25Ω [10 Ω+25Ω] RTH = 32Ω | Rtotal= 10Ω + (25Ω//25Ω) = 10Ω + [(25Ω × 25Ω0] [(25Ω + 25Ω)] = 10Ω + 12.5ΩRtotal = 22.5 ΩItotal = 12V/22.5Ω= 0.53 AIsc= 0.53A × [ 25Ω] [25Ω + 25Ω] = 0.53A × 0.5Ω = 0.265 ARN = 25Ω + (25Ω//10Ω) = 25Ω + [25 × 10] [25 + 10]RN = 32.14 ΩNorthon equivalent circuit:Current in 15Ω resistor:IL = 0.53 × [ 32.14Ω] [32.14Ω + 15Ω] = 1.21 A |

1.1 TITLE OF EXPERIMENT
Thevenin Theorem
1.2 OBJECTIVE * To calculate the parameters of AC equivalent circuits transformation theorems * To apply theory techniques to the solution of AC circuit problems.
1.3 THEORY
Thevenin’s Theorem states any linear two port network can be replaced by a single voltage source with series impedance. While the theorem is applicable to any number of voltage and current sources, this exercise will only examine single source circuits for the sake of simplicity. The Thevenin voltage is the open circuit output voltage. This may be determined experimentally by isolating the portion to be Thevenized and simply placing an oscilloscope at its outputs terminals. The Thevenin impedance is found by replacing all sources with their internal impedance and then applying appropriate series-parallel impedance simplification rules. If an impedance meter is available, an easy method of doing this in the lab is to replace the sources with appropriate impedance values and apply the impedance meter to the output terminals of the circuit portion under investigation.
1.4 EQUIPMENTS * Oscilloscope * Multimeters * Function generator * Resistors : 51Ω, 1.0KΩ, 1.2KΩ * Inductor : 10mH * Capacitor : 1µF, 4.7µF
1.5 PROCEDURES

Figure 1 : Circuit 1 Figure 2 :Thevenin Equivalent Circuit

Figure 3 : Circuit 2

1. For the circuit of figure 1, the voltage across the 1kΩ load was calculated using R1=1.2kΩ, R2=2.2kΩ, and C=4.7µF, with a 1kHz source. The value in table 1 was recorded. The expected thevenin voltage and impedance was also calculated. These values are then recorded in table 2. 2. The circuit of figure 1 was built using R1=1.2kΩ, R2=2.2kΩ, C=4.7µF and Rload=1kΩ. The generator was set to a 1kHz sine wave at 1Vpeak. The bandwidth limit of the oscilloscope was made sure was engaged. This reduced the signal noise and made for more accurate readings. The measured load voltage are then recorded in table 1. 3. The load was removed and the unload output voltage was measured. This is the experimental Thevenin voltage. It was recorded in table 2. 4. The internal resistance of function generator was measured. R generator = _________ 5. The voltage source was replaced with a resistor of R generator to represent its internal impedance. The impedance meter was set to 1kHz and the resulting impedance was measured at the open load terminals. This is the experimental Thevenin impedance. These values are recorded in table 2 and compared with the theoretical values. 6. The decade resistance box and capacitor was used, and the thevenin equivalent circuit of figure 2 was built and the load resistor of 1kΩwas applied. The load voltage was measured and recorded in table 1. The values of the original circuit and the thevenin circuit was compared and the deviation between them are calculated. 7. Steps 1 to 5 was repeated to verify that Thevenin Theorem also works on inductive source and complex loads. Figure 3 was used with R1=1.2kΩ, R2=2.2kΩ, L=0.8Mh, Rload=1kΩ with Cload=1µF. The generator was set tot a 10kHz sine wave at 1Vpeak. The results was tabulated in Tables 3 and 4.
1.6 RESULTS AND ANALYSIS AC equivalent circuits AC circuit | AC equivalent circuit | C
47µF
R1
1.2kΩ
R2
2.2kΩ

Vin

Rload
1kΩ
| RTH

Rload

VTH

| AC circuit | AC equivalent circuit | Cload
1µF
R1
1.5kΩ
R2
2.2kΩ
L1
0.8mH

Vin

Rload
1kΩ
| RTH

Rload

VTH

| CIRCUIT 1 | CIRCUIT 2 | RTH = [(1.2k + j0)//(2.2k + j0)]+ (0-j4.7µ) = [(1.2k +j0)(2.2k + j0)] + (0-j4.7µ) [(1.2k +j0)+(2.2k + j0)] = [(2640k + j0)] + (0-j4.7µ) [ (3400 + j0)] = (776.42 + j0) + (0-j4.7µ) = (776.47 – j4.7µ) = 776.472+4.7µF2 = 776.47 ΩXC = 1/2π × 1k × 4.7µ = 7.38 Ω.Z = RTH + XC = 776.47 - j 7.38 Ω Z = RTH2 -XC2 = (776.472)+(-7.382) = 776.43ΩVTH = [ (2.2k + j0)] × (1 + j0) [(1.2k + j0) + (2.2k + j0)] = [(2.2k + j0)] × (1 + j0) [(3400 + j0)] = (0.647 + j0) × (1 + j0) = (0.647 + j0) = 0.647^2 + 0^2 = 0.647 V | RTH =[(1.2k + j0)//(2.2k + j0) + (0 + j0.8m) = [(1.2k +j0)(2.2k + j0)] + (0 + j0.8m) [(1.2k +j0)+(2.2k + j0)] = [(2640k + j0)] + (0 + j0.8m) [ (3400 + j0)] = (776.47 + j0) + (0 + j0.8m) = (776.47 + j0.8m) = 776.472+(0.8mH)^² = 776.47 ΩXL = 2πFL = 2π (10)(0.8mH) = 0.05 ΩZ = RTH + XL = 776.47 + 0.05 Ω = 776.52 ΩVTH = [ (2.2k + j0)] × (1 + j0) [(1.2k + j0) + (2.2k + j0)] = [(2.2k + j0)] × (1 + j0) [(3400 + j0)] = (0.647 + j0) × (1 + j0) = (0.647 + j0) = 0.647^2 + 0^2 = 0.647 V |
Thevenin Theorem Calculations

Data of experiment Vload theory | | Vload original | 26.384µV | Vloadthevenin | 58.241µV | % Deviation | |
Table 1 | Theory | Experimental | % Deviation | V thevenin | | 57.141µV | | Z thevenin | | 1.035mV | |
Table 2 Vload theory | | Vload original | | Vloadthevenin | | % Deviation | |
Table 3 | Theory | Experimental | % Deviation | V thevenin | | | | Z thevenin | | | |
Table 4

1.7 CONCLUSION AND DISCUSSION 1. Explain the AC version of Thevenin’s Theorem compare with the DC version.
In the AC version of Thevenin Theorem we need to use imaginary numbers to calculate it’s RTH and VTH while in DC version we do not need to use imaginary numbers. Another difference between AC version and DC version is that in AC version we sometimes do include impedance instead of only resistance as in DC version. 2. Discuss the condition when the source frequency of Thevenin equivalent circuit was changed.
When the source frequency of Thevenin equivalent circuit was changed, the other value of RTH and VTH of the equivalent circuit will also changed based on the source frequency. 3. Identify other theorem that can be applied in this experiment other than Thevenin Theorem. Explain briefly.
Another theorem than can be applied in this experiment is Norton Theorem. This is because Norton Theorem also does the same job as Thevenin Theorem as they help simplify a circuit into an equivalent circuit. Besides Norton Theorem, Delta and Wye circuits are also similar to Thevenin Theorem. In a 3 phase power circuit, three phase source and loads can be connected in 2 ways which are delta and wye connections. This helps simplify the circuits into an equivalent wye and delta circuits so that it is easier to be analyze.