Question

State Thevenin’s theorem and give a proof. Apply this theorem to calculate the current through the 4 Ω resistor of the circuit of Fig.

Answer 1

As shown in Fig., 4 Ω resistance has been removed thereby open-circuiting the terminals A and B. We will now find V_AB and R_AB which will give us V_th and R_th respectively. The potential drop across 5Ω resistor can be found with the help of voltage-divider rule. Its value is 

= 15 × 5/(5 + 10) = 5V.

For finding V_AB, we will go from point B to point A in the clockwise direction and find the algebraic sum of the voltages met on the way. 

∴ V_AB = − 6 + 5 = − 1 V. 

It means that point A is negative with respect to point E, or point B is at a higher potential than point A by one volt. 

In Fig., the two voltage source have been short circuited. The resistance of the network as viewed from points A and B is the same as viewed from points A and C.

 

∴ R_AB = R_AC = 5 || 10 = 10/3 Ω 

Thevenin’s equivalent source is shown in Fig. in which 4 Ω resistor has been joined back across terminals A and B. Polarity of the voltage source is worth nothing.

∴  I = 1/((10/3) + 4) = 3/22 = 0.136A