1. Construct the circuit shown in Figure 1. Calculate all currents and voltage d

ID: 1811033 • Letter: 1

Question

1. Construct the circuit shown in Figure 1. Calculate all currents and voltage drops by using the Superposition Theorem.

2. Measure and record the voltage drops across and current through all resistors.

3. Remove the 10 volt source and replace it with a short circuit. Repeat step 2.

4. Reinstate the 10 volt source and replace the 5 volt source with a short circuit. Repeat step 2.

5. Construct the circuit in Figure 2. Calculate Eth and Rth for the circuit at the output terminals, A and B

6. Place RL at points A and B. Measure the load voltages for RL = 100, 330, 1 k, and 10 k ohm

7. Construct the Thevenin equivalent circuit and repeat step 6.

1)using super position therom

we shall find individual responces then add the responces

drops due to E1short E2

finding current I = E1/(680+330||470)=10/873.85 = 11.44 mA

current in 330 ohm resistor is = I*470/(470+330) = 6.72 mA

current in 470 ohm resistor is = I*330/(470+330) = 4.72 mA

drop across 680 ohm is = 680*I= 7.792 V

drop across 330 ohm is = 370*6.72 m= 2.21 V

drop across 470 ohm is = 470*4.72 m= 2.21 V

drops due to E2 short E1

finding current I = E2/(470+330||680)=5/692.17 = 7.22mA

current in 330 ohm resistor is = I*680/(680+330) = 4.86 mA

current in 470 ohm resistor is = I*330/(680+330) = 2.36 mA

drop across 470 ohm is = 470*I= 3.393 V

drop across 330 ohm is = 370*4.86 m= 1.6 V

drop across 680 ohm is = 680*2.36 m= 1.6 V

using superposition principle total responce is sum of individual responces

drop across 680 = drop due to E1 + drop due to E2

7.792 V +1.6 V =9.39 V

drop across 470 = drop due to E1 + drop due to E2

2.21V +3.93 V =6.14 V

drop across 330 = drop due to E1 + drop due to E2

2.21 V +1.6 V =3.81 V

upto question no 4 is covered in the above solution

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