Circuit Theory/Lab4.5.1
Example, find the thevenin equivalent of this circuit, treating R7 as the load.
- Simulate the circuit, displaying load voltage and current as the load is swept through a range of resistance values
- Simulate the thevenin equivalent circuit and again sweep the load voltage and current through a range of resistance values
Finding Thevenin Voltage[edit | edit source]
Open the load (resistor R7), and find the voltage across it's terminals.
R5 and R6 are dangling and can be removed.
Vth = VA - VB
VA = 2-VR1 = 2 - (2-5)*2.2/(2.2+4.7) .. voltage divider
VA = 2.9565
VB = 5-VR3 = 5 - 5*6.8/(6.8+6.8) .. voltage divider
VB = 2.5 volts
Vth = 2.9565 - 2.5 = 0.4565 volts
Can check with this simulation.
Finding Thevenin Resistance[edit | edit source]
Remove the load, zero the sources.
Redraw up and down so the parallel/serial relationships between the resistors are obvious.
Finding Norton Current[edit | edit source]
IN = Vth/Rth = 0.4565/7.3986 = 0.0617 amp
Simulating the original circuit[edit | edit source]
In the simulation, can see the computed Norton's current when the load is 0 ohms.
Can see the computed Thevenin voltage when the load is around 20 ohms which approximates an open.
Comparing with the Thevenin Equivalent[edit | edit source]
In this simulation, can see the same values, except this time the load voltage is relative to ground, so don't have to look at a drop or differences between two voltages as with the original circuit simulation.
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Sweep simulation of original circuit for R7 values of 47 ranging from 0 to 20
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The thevenin equivalent with the R7 load in place
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Sweep simulation of the thevenin equivalent circuit for R7 values ranging from 0 to 20 ohms