|
RC1
|
RC2
|
RC3
|
RL1
|
VS1
|
VS2
|
VL2
|
|
100 Ω
|
39 Ω
|
39 Ω
|
680 Ω
|
9 V
|
9 V
|
8 V
|
STEP 1) Applying thevenin's theorem to calculate for the unknown RL2.
In order to accomplish the objective, the thevenin equivalents must be determined. We started off by finding the thevenin voltage, Vth.
The branch with L2 was taken off to form an open circuit across the terminals.
The voltage across the terminals of the open circuit would then be the thevenin voltage.
Assuming current i1, i2, i3 were all comming out from the node, KCL concludes that i1+i2+i3=0.
The thevenin voltage was approximately 8.64 V.
Now, we solve for the Norton current and the thevenin resistance.
This time, the branch is short circuited, and by definition, the short circuited current Isc is the Norton current.
Another nodal analysis was performed to calculated Vy at the same node. Knowing that Vy = 5.11 V, we divide Vy by RC3 according to Ohm's Law to find the short citcuited current Isc=0.131A.
Therefore, according to Ohm's law again, the thevenin resistance is equal to thevenin voltage divided by Norton Current.
Rth = Vth/IN = 65.5 Ω
Finally, we transformd the circuit to simplify calculation and solved for RL2 min.
The circuit was simplified to a voltage source in series with 2 resistors; a simple voltage divider rule will yield the answer.
RL2 min = 870.2 Ω
STEP 2) Build the circuits and simulate the experiment to confirm the calcuated values.
We first simulate the more simple thevenin equivalent circuit.
The components used in the circuits were:
|
Component
|
Nominal Value
|
Measured Value
|
Power or Current Rating
|
|
R_Th
|
66 Ω
|
65.8 Ω
|
1 W
|
|
R_L2,min
|
876 Ω
|
873 Ω
|
1 W
|
|
V_Th
|
9 V
|
9.21 V
|
2 A
|
We measured the voltage across Load, L2, when the load had the same resistance as the thevenin resistance, as well as, when the circuit is shorted.
The data collected were as the following:
|
Config
|
Theoretical Value
|
Measured Value
|
Percent Error
|
|
RL2 = RL2,min
|
VL2 = 8 V
|
8.5V
|
6.25 %
|
|
RL2 = ∞ Ω
|
VL2= 8.6 V
|
9.17 V
|
6.63 %
|
Now, we remove the thevenin equivalent circuit with the original circuit.
Measuring the voltage across L2
The circuit had these components:
|
Component
|
Nominal Value
|
Measured Value
|
Power or Current Rating
|
|
R_C1
|
100 Ω
|
99.1 Ω
|
1/8 W
|
|
R_C2
|
39 Ω
|
38.1 Ω
|
1/8 W
|
|
R_C3
|
39 Ω
|
38.6 V
|
1/8 W
|
|
R_L1
|
680 Ω
|
667 Ω
|
1/8 W
|
|
V_S1
|
9 V
|
9.26 V
|
2 A
|
|
V_S2
|
9 V
|
9.21 V
|
2 A
|
We disconnect L2 from the circuit, and measure the voltage acorss L2 at its terminals as we had once been done for the thevenin equivalent circuit.
|
Config
|
Theoretical Value
|
Measured Value
|
Percent Error
|
|
R_L2 = R_L2,min
|
V_Load2 = 8 V
|
8.20 V
|
2.5 %
|
|
R_L2 = ∞ Ω
|
V_Load2 = 8.6 V
|
8.84 V
|
2.79 %
|
Last Step) Observe the cnage in power devliverd to the thevenin resistaor as it resistace changes.
Theoretical circuit analysis:
According to Joule's heating, Power = (I^2)*R = (V^2)/R
The power going into R_L2 = R_Th is V^2/R = 0.280 W
Calculating the first row is 2.97^2/(0.5*R_Th) = 0.267 W
The voltage acorss the adjustable L2 will be recorded as the resistance of L2 changes.
Config
|
V_Load 2
|
P_Load 2
|
R_L2 = 0.5R_Th
|
2.97 V
|
0.267 W
|
R_L2 = R_Th
| ||
R_L2 = 2R_Th
|
5.89 V
|
0.264 W
|
Thevenin Equivalent did work!
Instead of recalculating the whole circuit again and again after adding more components, networks of circuits can be simplified into a thevenin power source and a thevenin resistor to simplify calculations!!!!
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