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Question:
Published on: 25 April, 2024

State and explain Thevenins theorem. Find the Thevenins equivalent of the circuit of Fig.18(a) as shown at terminal XY.

Fig. 18(a)

Answer:

Thevenin’s theorem states that any two terminal bilateral linear d.c circuits can be replaced by an equivalent circuit consisting of a voltage source and a series resistor.

Explanation:

Let us consider simple d.c. circuit as shown in Fig.17 (a), we are to find IL by Thevenin's theorem. In order to find the equivalent voltage source, rL is removed and Vo.c is calculated as shown in Fig. 17(b).

                     

                    (a)                                                                   (b)

                      

                   (c)                                                             (d)

Fig. 17(a) Simplified d.c. circuit, (b) Finding of Vo.c, Finding of Rth and (d) Finding of IL forming Thevenin’s equivalent circuit.

Next, to find the internal resistance of the network (Thjevenin’s resistance or equivalent resistance) in series with Vo.c, the voltage source is removed (de active) by a short circuit (as the source does not have any internal resistance) as shown in Fig .17(c).

 

                           

                       (b)                                                                       (c)

(d)

Fig. 18 (b) Finding of Vo.c, (c) Finding of Rth and (d) Finding of IL forming Thevenin’s equivalent circuit.

The current I1 through the 20 Ω resistance due to the current source 1 Amp and voltage source 32 V is removed.

I1=1×8/30=8/30 Amp.

The current I2 through the 20 Ω resistance due to the voltage source 32 V and current source 1 Amp. is removed.

I2=32/30 Amp.

The total current through the resistance 20 Ω is I=I1-I2=0.27-1.07=-0.8 Amp.

The current is following upward direction through the 20 Ω resistor.

The voltage drop between terminal X and Y is the open circuit voltage

Vo.c=32-20×0.8=32-16=16 V

Thevenin’s equivalent resistance Rth=(10×20)/(10+20)=6.67 Ω

The load current IL=Vo.c/(Rth+R)=16/(6.67+R) Amp.

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