Thevenin's Theorem
In this lab we use the are given the circuit above with two independent sources (5 V and 2 V) and 6 resistors along with a variable load resistor R_L. Thevenin's theorem makes it possible to create an equivalent circuit consisting of an equivalent voltage source V_th in series with an equivalent resistor R_th. The load, R_L can be changed many times and calculations become simpler.
We first begin by finding the equivalent Thevenin resistance, R_th (calculation in purple). We remove all independent voltage sources (shorting them), remove the load resistor R_L, and measure the resistance from nodes a and b. We obtain an equivalent resistance of R_th = 7.1986 kΩ.
We now calculate for the open circuit voltage across nodes a and b (calculation in red). The open circuit voltage is the Thevenin equivalent voltage, V_th. This time we factor in the voltage sources but leave the load resistor out (R_L). We use mesh analysis to find our two mesh currents i_1 and i_2. There is no current flowing through the 1 kΩ and 1.3 kΩ resistors because a-b is an open circuit. Using the mesh currents found, we calculate the voltage drops across the 4.7 kΩ and 6.8 kΩ resistors. The sum of these voltage drops equals the voltage drop from a to b which is V_th. V_th = V_4.7 + V_6.8 = -2.04 + 2.50 = 0.45 V.
We wire the circuit without the voltage sources and measure the Thevanin equivalent resistance, R_th = 7.17 kΩ (0.4% error).
We add the voltage supplies and measure the voltage drops across the 4.7 kΩ and 6.8 kΩ resistors. The sum of of these voltage drops is our equivalent Thevanin voltage. V_th = 2.48 - 2.02 = 0.46 V (2.2% error).
We now diagram our Thevanin equivalent circuit with its equivalent voltage source V_th = 0.458 and equivalent Thevanin resistance R_th = 7.1986 kΩ. Providing the 458 mV source is not a problem using WaveForms. However, finding a 7.20 kΩ resistor is a challenge and ordering one can be costly. Instead, we use a 10k potentiometer and set it to have a resistance of 7.20 kΩ. We now wire the circuit with 458 mV source (red probe), potentiometer as the Thevalin resistor, load resistor R_L connected to ground (black probe).
We now measure the load voltages across the 1 kΩ (left) and 6.8 kΩ (right) resistors. We obtain load voltages for the 1 kΩ and 6.8 kΩ resistors of V_L(1kΩ) = 54 mV (55.8 mV theoretical, 3.2% error) and V_L(6.8 kΩ) = 216 mV (225 mV theoretical, 4.0% error).
If we wanted to repeat the calculations for many load resistances, it is convenient to use another potentiometer. We used a 10k potentiometer (9.2k measured) in place of the load resistor. We then used two multimeters to measure the load resistance and load voltage simultaneously. However, we measure the resistance 'on the other side' of the pot. In this case, we need to subtract the measured resistance from the pot resistance from side to side. So we subtract the multimeter reading from 9.2 kΩ to obtain the load resistance provided by the potentiometer.
We first begin by finding the equivalent Thevenin resistance, R_th (calculation in purple). We remove all independent voltage sources (shorting them), remove the load resistor R_L, and measure the resistance from nodes a and b. We obtain an equivalent resistance of R_th = 7.1986 kΩ.
We now calculate for the open circuit voltage across nodes a and b (calculation in red). The open circuit voltage is the Thevenin equivalent voltage, V_th. This time we factor in the voltage sources but leave the load resistor out (R_L). We use mesh analysis to find our two mesh currents i_1 and i_2. There is no current flowing through the 1 kΩ and 1.3 kΩ resistors because a-b is an open circuit. Using the mesh currents found, we calculate the voltage drops across the 4.7 kΩ and 6.8 kΩ resistors. The sum of these voltage drops equals the voltage drop from a to b which is V_th. V_th = V_4.7 + V_6.8 = -2.04 + 2.50 = 0.45 V.We wire the circuit without the voltage sources and measure the Thevanin equivalent resistance, R_th = 7.17 kΩ (0.4% error).
We add the voltage supplies and measure the voltage drops across the 4.7 kΩ and 6.8 kΩ resistors. The sum of of these voltage drops is our equivalent Thevanin voltage. V_th = 2.48 - 2.02 = 0.46 V (2.2% error).
We now apply a load resistor R_L of 1 kΩ (0.994 kΩ measured) and 6.8 kΩ (6.63 kΩ measured) resistances, one at a time. We then measure the load voltage drops V_L across these resistors.
First we apply the 1 kΩ resistor and measure a V_L of 54.4 mV (55.9 mV theoretical, 2.7% error). We then apply the 6.8 kΩ resistor and measure a V_L of 214 mV (225 mV theoretical, 4.9% error).
The calculated results were closely to their corresponding theoretical results.
We now diagram our Thevanin equivalent circuit with its equivalent voltage source V_th = 0.458 and equivalent Thevanin resistance R_th = 7.1986 kΩ. Providing the 458 mV source is not a problem using WaveForms. However, finding a 7.20 kΩ resistor is a challenge and ordering one can be costly. Instead, we use a 10k potentiometer and set it to have a resistance of 7.20 kΩ. We now wire the circuit with 458 mV source (red probe), potentiometer as the Thevalin resistor, load resistor R_L connected to ground (black probe).
We obtain the following data where the first two columns are the readings from the multimeter. The R_adjusted column is 9.2 kΩ - R_meter. Power is just calculated as (V_meter)^2/R_adjusted. Units (not shown in data) are in V, kΩ, kΩ, and mW across the columns from left to right.
We now can plot a graph for power vs. resistance. We can see that the power increases as the load resistance increases until it reaches a maximum and then starts to fall a bit. The maximum is 6.97 mW, which occurs when our load resistance is 7.26 kΩ. This corresponds closely to our Thevenin resistance of 7.20 kΩ. Maximum power will occur when the load resistance is equal to Thevenin resistance.
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