Figure 1: Thevenin's Theorem Circuit |
All of resistors are measured by a digital Ohmmeter, and those are written by red color (Fig 1).
Figure 2: Calculation of Thevenin's Voltage |
According to measured resistors in the above circuit, Thevenin's Voltage is calculated.
(2.17 + 4.58)*I1 = 2-5
(6.65 + 6.69)*I2 = -5 , I1 = - 0.444 A , I2 = - 0.375 A
Vc - 4.58*I1 + 6.69*I2 = Vd , Vc - Vd = Vth = 4.58*(-0.444) - 6.69*(-0.375) = 0.475 v
Theoretical Thevenin's voltage is 0.475 v , and experimental Thevenin's voltage is 0.47 v (Fig 1).
Percent Thevenin's Voltage = [(0.475-0.47)/0.475]*100 = 1.1%
Figure 3: Thevenin's Voltage is measured 0.47 volts |
When the load resistor changes, its drop voltage will measured by a digital voltmeter, and it records in order to determine drop power of the load resistor.
Load Resistor (Ω)
|
Drop Load Voltage (v)
|
Drop Load Power (W)
|
2.17E+03
|
0.108
|
5.38E-06
|
3.53E+03
|
0.153
|
6.63E-06
|
4.62E+03
|
0.183
|
7.25E-06
|
5.61E+03
|
0.204
|
7.42E-06
|
6.80E+03
|
0.227
|
7.58E-06
|
7.40E+03
|
0.238
|
7.65E-06
|
8.02E+03
|
0.247
|
7.61E-06
|
9.80E+03
|
0.27
|
7.44E-06
|
1.78E+04
|
0.334
|
6.27E-06
|
When P(load) is maximum?
At now, circuit has a voltage source equal Thevenin's voltage, and a resistor equal Thevenin's resistor. In this situation, when R(load) is equal with Rth, P(load) will be maximum.
P(load) = [(Vload)^2] / RL , V(load) = [RL / (RL+Rth)]*Vth → P(load) = (RL)*[Vth / (RL+Rth)]^2
Derivative of P(load) relative to RL equal zero is [(RL+Rth)^2] - (RL)*[2*(RL+Rth)] = 0
(RL+Rth) - 2*RL = 0 , RL = Rth
According to the above table, P(load) is maximum when RL is equal 7.4 k.
This is the formula P(load) = (RL)*[0.47/(7400 + RL)]^2 that graph figure 4 follows.
Figure 4: Load Power vs Load Resistor |
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