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Unformatted text preview: – Thevenin Equivalent Resistance – The current through a 510Ω resistor across A and B in the equivalent circuit Table 1. The various values before the Thevenin Equivalent circuit was built Value Calculated Measured Percent Error 9.17 V 9.18 V 0.11% 15.75 mA 14.41 mA 8.51% 635.14 Ω 629 Ω 0.94% 8.734 mA 8.02 mA 8.90% Question 1. With the 510Ω load resistor in place, how much power does the entire Thevenin’s equivalent circuit consume? With the load resistor, the current was measured to be 8.02mA. Coupled with the voltage, 9.18V, we can find the power thus so: II. Maximum Power Transfer In this section of the lab, we had to connect the load resistance across the terminals A and B in our equivalent circuit. We then measured the values in Table 3 and plotted them in Figure 3. Table 3. The power measurements Resistance () Voltage () Current () Power () 510 9.17 8.02 32.803 560 9.17 7.64 32.687 620 9.17 7.3 33.040 680 9.17 6.97 33.035 750 9.17 6.64 33.067 Figure 3. The plot of power versus resitance, showing a slight ‘bump’ around the value III. Simulations Finally, each circuit is simulated here, using Multisim . Figure 4. The initial circuit, with a multimeter reading Figure 5. The initial circuit, with a multimeter reading With these two values, we can calculate the equivalent resistance using Ohm’s Law: Figure 6. The Thevenin equivalent circuit with a multimeter reading Figure 7. The Thevenin equivalent circuit with a multimeter reading Figure 8. The Thevenin equivalent circuit with a 636Ω resistor across terminal AB, with a Figure 8....
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 Fall '11
 Jafarzadeh
 Volt, Resistor, Thévenin's theorem, Voltage source, Norton's theorem, equivalent circuit

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