When the circuit have many sources and multiple loads, then we could calculate the thevenin to make the problem easier.
Thursday, March 31, 2011
Laboratory #6 Thevenin Equivalents
This lab we did on this Monday that used thevenin method to make the complicated circuits look easier.
When the circuit have many sources and multiple loads, then we could calculate the thevenin to make the problem easier.
Before the lab began, we had the set up number for resistors and the voltage of the sources. They were Rc1= 100 ohms, Rc2=Rc3=39 ohms, RL1=680 ohms and Vs1=Vs2= 9V. Then, we could calculate theoretical value for the Vx by using the nodal analysis. (9-Vx)/39+(9-Vx)/100=Vx/680, so the Vx=8.6433V. The second part was to calculate the Vy by connecting the two points. (Vs2-Vy)/Rc2+(Vs1-Vy)/Rc1=Vy/RL1+Vy/Rc3, the Vy=5.11V. Furthermore, we got the Isc=Vy/Rc3=0.131A and Rth= 65.97ohms. If the ab point carried 8v; then, the resistor should be (8.64-8)/66=8/X, so the resistor X=825 ohms. I would be the same with Isc and V was the same with Vth=8.64 V. After we calculated all the theoretical values, we did set up the circuit like the image, and we almost got perfect numbers. We were used Rth=66.6 ohms and RL2=825 ohms and Vth=8.64V, all the numbers were real close to the nominal value. The value we measured were 7.98V for Vload 2 when RL2=RL2, min, the percentage error was 0.25%, and when RL2= infinity, the Vload 2=8.64 the measured value was 8.63V so the percentage error was 0.11%. I thought this experiment was perfect.
When the circuit have many sources and multiple loads, then we could calculate the thevenin to make the problem easier.
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