# 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 redrawing thevenin resistance network to make calculation easier

## Finding Thevenin Voltage

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

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 Current

IN = Vth/Rth = 0.4565/7.3986 = 0.0617 amp

## Simulating the original circuit

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

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.