7.07 Using the equivalent circuits.

We will connect a 10 Ohm load resistor to the original circuit and to the Norton and Thevenin equivalent circuits. Then for each circuit, we will calculate the voltage across the load and the current through it, to show that the original circuits and both of the equivalent circuits predict the same values.

Original circuit.

Original circuit

R_T = (R₁ x R_L)/ (R₁ + R_L) + (R₂ x R₂)/ (R₂ + R₃) .
R_T = 8.857 Ohms .
I_T = E/R_T .
I_T = 1.129 A .
Using the current divider rule I_L = I_T x R₁/(R₁ + R_L).
I_L = 0.32A .
V_L = I_L x R_L
V_L = 3.2 V .

Thevenin equivalent circuit.

Thevenin circuit

Using the voltage divider rule V_L = E x R_L/(r + R_L) .
V_L = 3.2 V .
I_L = V_L/R_L .
I_L = 0.32A .

Norton equivalent circuit.

Norton circuit

Using the current divider rule I_L = I_SC x r/(r + R_L) .
I_L = 0.32A .
V_L = I_L x R_L .
V_L = 3.2 V .

As you can see, both the Thevenin and Norton equivalent circuits, predict the same values of I_L and V_L as the original circuit. It is also clear that there are fewer calculations required for the equivalent circuits than for the original circuit, even though the original circuit is not itself very complicated.