# 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 VoltageEdit

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

R_{5} and R_{6} 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.

## 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.