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Step of 8 1.015P Thevenin equivalent of a circuit is a circuit which consists of a voltage source and a series resistance where series resistance is the equivalent resistance of the circuit where voltage source has a voltage equal to the voltage across the load terminal. Refer to the Figure P1.15 in the textbook. Step of 8 Consider the nodes 3 and 4 according to voltage divide rule voltage across the node 3 is twice the voltage across the node 4. Let voltage across the node 4 is and hence voltage across the node 3 is Since, two equal resistors are connected to node 3(there will be equal drop of voltage at each resistor). Similarly between the nodes 2 and 3, according to voltage division rule voltage across the node 2 is twice the voltage across the node 3. Hence, voltage across the node 2 is, Step of 8 Consider the nodes between 1 and 2, according to voltage division rule voltage across thje node 1 is twice the voltage across the node 2. Thus, voltage across the node 1 is, = Step of 8 The voltage across the node 1 is 10 10 8 The voltage of node 4 is equivalent to the Thevenin's voltage. Therefore, thevenin's voltage is 1.25 V Step of 8 Determination of Thevenin's resistance as follows: Short circuit the voltage sources and open circuit the current sources to find the equivalent resistance across the load terminal. The resistors between nodes 1 and 2 are parallel to each other and hence equivalent resistance is, Step of 8 and 20 kΩ between nodes 2 and 3 are in series. Then equivalent resistance is, R2 kΩ and 20 kΩ between nodes 3 and ground are in parallel then equivalent resistance is = Step of 8 R3 and 20 kΩ are in series and their equivalent resistance is, 32 kΩ and 20 kΩ between node 4 and ground are in parallel and their equivalent resistance is simply thevenin's equivalent resistance. 32x20 = 32+20 Therefore, the Thevenin's resistance is Consider the Load resistance of the circuit is 3 kΩ Thevenin equivalent of a circuit is shown in Figure 1. Figure Step of 8 Determine the current is flowing through load resistance 3 kΩ. 1.25 V II Hence, the current that flows through a load resistance of 3 is