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

the original circuit, R7 is the load, find thevenin equivalent of the rest of the circuit

Finding Vth using ohm's law

thevenin resistance is calculated by zeroing the sources and opening the load

redrawing thevenin resistance network to make calculation easier

Finding Thevenin VoltageEdit

Open the load (resistor R7), and find the voltage across it's terminals.

R₅ and R₆ are dangling and can be removed.

V_th = V_A - V_B

V_A = 2-V_R1 = 2 - (2-5)*2.2/(2.2+4.7) .. voltage divider

V_A = 2.9565

V_B = 5-V_R3 = 5 - 5*6.8/(6.8+6.8) .. voltage divider

V_B = 2.5 volts

V_th = 2.9565 - 2.5 = 0.4565 volts

Can check with this simulation.

Finding Thevenin ResistanceEdit

Remove the load, zero the sources.

Redraw up and down so the parallel/serial relationships between the resistors are obvious.

$R_{th}=1+{\frac {1}{{\frac {1}{2.2}}+{\frac {1}{4.7}}}}+{\frac {1}{{\frac {1}{6.8}}+{\frac {1}{6.8}}}}+1.5$
$R_{th}=7.3986$

Finding Norton CurrentEdit

I_N = V_th/R_th = 0.4565/7.3986 = 0.0617 amp

Simulating the original circuitEdit

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 EquivalentEdit

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.

Sweep simulation of original circuit for R7 values of 47 ranging from 0 to 20
The thevenin equivalent with the R7 load in place
Sweep simulation of the thevenin equivalent circuit for R7 values ranging from 0 to 20 ohms