05.1 Thievenin Equivalent Circuit Practice Problems

 

THEVENIN EQUIVALENT CIRCUIT |   P RACTICE PROBLEMS    

Complete the following problems to reinforce your understanding of the concept covered in this module.  

PROBLEM 1:  

What is the Thevenin equivalent at terminal a-b for the circuit shown below? Find the total current in the circuit.

 

PROBLEM 2:

What is the Thevenin equivalent at terminal a-b for the circuit below? Find the Thevenin Equivalent Resistance and Thevenin Equivalent Voltage in the circuit.

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PROBLEM 3:

What are the Thevenin equivalent resistance and voltage between terminals A and B?

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THEVENIN EQUIVALENT CIRCUIT |   SOLUTIONS  

 

SOLUTION 1:

For this circuit, we have only one voltage source unlike the concept example that had two voltage sources. 1. Remove the load resistance, R L , from the circuit. In this example, it is the 3Ω . 2. Open the current sources and short the voltage sources. 3. Calculate the equivalent resistance looking from the load terminal. The 4Ω resistor is in parallel with the 2Ω resistor, so the equivalent resistance is calculated as:

R TH =

R1 × R2 4 × 2 = = 1.33Ω R 1 +R 2 4 + 2

4. Calculate the source voltage by reconnecting the voltage source in the circuit. Since there is one voltage source VS = 8V Calculate the source current flowing in the circuit using Ohm’s Law. Plug in the previously calculated source voltage and the equivalent resistance with the source voltage connected:

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VTH =

 

R1 V. R1 + R2 S

This formula is not supplied in the Reference Handbook. 5. Calculate the voltage drop across the 4Ω resistor with the 8 V voltage source:

(

)

VTH   = VS  –   R 4Ω  x IS   

(

)

VTH   =  8V  –   4" x 1.33amps   =   2.68 Volts

The voltage drop across the 2Ω resistor will be the same since the resistors are in parallel. The Thevenin Voltage is the same across each resistor:

VTH = 2.68Volts 6. We can now calculate the total current flowing through the circuit using Ohm’s Law expressed as:

IT =

IT =

VTH R TH + R L

2.68v = 0.619 amps 1.33!+3!

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SOLUTION 2:

This problem is an easy Thevenin Equivalent Circuit problem as there is only one voltage source, and there are only two resistors, which are in parallel. To calculate the Thevenin Equivalent Voltage, we can use the voltage divider rule: voltage source, and there are only two resistors, which are in parallel.

VTH = VTh = 

R1 V , where VS is the source voltage R1 + R2 S

(

)( )

10 15 = 3.75 V 10 + 30

To calculate the Thevenin Equivalent Resistance, we solve for the equivalent resistance of the two resistors in parallel:

R TH =

R 1R 2 R1 + R2

R TH =

(10Ω)(30Ω) = 7.5Ω (10Ω + 30Ω)

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SOLUTION 3: For this problem we will be solving for the Thevenin Equivalent Resistance and Thevenin Equivalent Voltage. The Thevenin resistance is the resistance between the two terminals when the voltage is source short-circuited. With the circuit shorted, the 16 Ω resistor is effectively bypassed, leaving the other two in a simple parallel arrangement.

R TH =

R 1R 2 R1 + R2

R TH =

(4Ω)(12Ω) = 3Ω (4Ω + 12Ω)

The Thevenin voltage is the voltage difference between the two terminals. In this case, it is equal to the voltage drop across the 12 Ω resistor. Apply Ohm’s Law once to find the current I12Ω , and then again to find the voltage across the resistor. 14|PREPINEER.COM    

 

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I12Ω =

V 60V = = 3.75A R 4Ω + 12Ω

VTH = I12Ω xR = (3.75A)(12Ω) = 45V

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