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Qmechanic ♦
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Thevenin's Equivalent Circuit  : Contradictory results by using 2 different methods?

Consider this Circuit which is to be converted to Equivalent Thevenin's Circuit.

Full Circuit

Now I tried Solving this using two methods and got contradictory result. We need $V_{Th}$ and $R_{Th}$

$$R_{Th}= 8\Omega + 8\Omega = 16\Omega$$

Now Finding $V_{Th}$: -

Method -$1$: - The current in $R_7$ branch is $20A$ and hence Voltage drop across the terminals $CD=$ $20A\times4\Omega = 80V$ and the direction of current points to fact that $V_C=80V$ and $V_D=0V$. Hence $V_{Th}=80+30=110V$

Thevenin's Circuit

Thevenin's Circuit

Method -$2$  : - Using Circuit Laws (like KVL/KCL/Superposition) to solve the circuit when Load is open to get $V_{Th}$

Solution

Solution

Solution

You can see that the polarity of the $V_{Th}$ is coming opposite in both. Why?

Thevenin's Equivalent Circuit  : Contradictory results by using 2 different methods?

Consider this Circuit which is to be converted to Equivalent Thevenin's Circuit.

Full Circuit

Now I tried Solving this using two methods and got contradictory result. We need $V_{Th}$ and $R_{Th}$

$$R_{Th}= 8\Omega + 8\Omega = 16\Omega$$

Now Finding $V_{Th}$: -

Method -$1$: - The current in $R_7$ branch is $20A$ and hence Voltage drop across the terminals $CD=$ $20A\times4\Omega = 80V$ and the direction of current points to fact that $V_C=80V$ and $V_D=0V$. Hence $V_{Th}=80+30=110V$

Thevenin's Circuit

Thevenin's Circuit

Method -$2$  : - Using Circuit Laws (like KVL/KCL/Superposition) to solve the circuit when Load is open to get $V_{Th}$

Solution

Solution

Solution

You can see that the polarity of the $V_{Th}$ is coming opposite in both. Why?

Thevenin's Equivalent Circuit: Contradictory results by using 2 different methods?

Consider this Circuit which is to be converted to Equivalent Thevenin's Circuit.

Full Circuit

Now I tried Solving this using two methods and got contradictory result. We need $V_{Th}$ and $R_{Th}$

$$R_{Th}= 8\Omega + 8\Omega = 16\Omega$$

Now Finding $V_{Th}$: -

Method -$1$: - The current in $R_7$ branch is $20A$ and hence Voltage drop across the terminals $CD=$ $20A\times4\Omega = 80V$ and the direction of current points to fact that $V_C=80V$ and $V_D=0V$. Hence $V_{Th}=80+30=110V$

Thevenin's Circuit

Thevenin's Circuit

Method -$2$: - Using Circuit Laws (like KVL/KCL/Superposition) to solve the circuit when Load is open to get $V_{Th}$

Solution

Solution

Solution

You can see that the polarity of the $V_{Th}$ is coming opposite in both. Why?

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Qmechanic ♦
Qmechanic ♦
  • 213.1k
  • 48
  • 590
  • 2.3k

Consider this Circuit which is to be converted to Equivalent Thevenin's Circuit.

Full Circuit

Now I tried Solving this using two methods and got contradictory result. We need $V_{Th}$ and $R_{Th}$

$$R_{Th}= 8\Omega + 8\Omega = 16\Omega$$

Now Finding $V_{Th}$  : -

Method -$1$  : - The current in $R_7$ branch is $20A$ and hence Voltage drop across the terminals $CD=$ $20A\times4\Omega = 80V$ and the direction of current points to fact that $V_C=80V$ and $V_D=0V$. Hence $V_{Th}=80+30=110V$

Thevenin's CircuitThevenin's Circuit

Thevenin's Circuit

Method -$2$ : - Using Circuit Laws (like KVL/KCL/Superposition) to solve the circuit when Load is open to get $V_{Th}$

SolutionSolutionSolution

Solution

Solution

You can see that the polarity of the $V_{Th}$ is coming opposite in both. Why?

Consider this Circuit which is to be converted to Equivalent Thevenin's Circuit.

Full Circuit

Now I tried Solving this using two methods and got contradictory result. We need $V_{Th}$ and $R_{Th}$

$$R_{Th}= 8\Omega + 8\Omega = 16\Omega$$

Now Finding $V_{Th}$  : -

Method -$1$  : - The current in $R_7$ branch is $20A$ and hence Voltage drop across the terminals $CD=$ $20A\times4\Omega = 80V$ and the direction of current points to fact that $V_C=80V$ and $V_D=0V$. Hence $V_{Th}=80+30=110V$

Thevenin's CircuitThevenin's Circuit

Method -$2$ : - Using Circuit Laws (like KVL/KCL/Superposition) to solve the circuit when Load is open to get $V_{Th}$

SolutionSolutionSolution

You can see that the polarity of the $V_{Th}$ is coming opposite in both. Why?

Consider this Circuit which is to be converted to Equivalent Thevenin's Circuit.

Full Circuit

Now I tried Solving this using two methods and got contradictory result. We need $V_{Th}$ and $R_{Th}$

$$R_{Th}= 8\Omega + 8\Omega = 16\Omega$$

Now Finding $V_{Th}$: -

Method -$1$: - The current in $R_7$ branch is $20A$ and hence Voltage drop across the terminals $CD=$ $20A\times4\Omega = 80V$ and the direction of current points to fact that $V_C=80V$ and $V_D=0V$. Hence $V_{Th}=80+30=110V$

Thevenin's Circuit

Thevenin's Circuit

Method -$2$ : - Using Circuit Laws (like KVL/KCL/Superposition) to solve the circuit when Load is open to get $V_{Th}$

Solution

Solution

Solution

You can see that the polarity of the $V_{Th}$ is coming opposite in both. Why?

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Mr. Cosmic_Quantum
Mr. Cosmic_Quantum
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Thevenin's Equivalent Circuit : Contradictory results by using 2 different methods?

Consider this Circuit which is to be converted to Equivalent Thevenin's Circuit.

Full Circuit

Now I tried Solving this using two methods and got contradictory result. We need $V_{Th}$ and $R_{Th}$

$$R_{Th}= 8\Omega + 8\Omega = 16\Omega$$

Now Finding $V_{Th}$ : -

Method -$1$ : - The current in $R_7$ branch is $20A$ and hence Voltage drop across the terminals $CD=$ $20A\times4\Omega = 80V$ and the direction of current points to fact that $V_C=80V$ and $V_D=0V$. Hence $V_{Th}=80+30=110V$

Thevenin's CircuitThevenin's Circuit

Method -$2$ : - Using Circuit Laws (like KVL/KCL/Superposition) to solve the circuit when Load is open to get $V_{Th}$

SolutionSolutionSolution

You can see that the polarity of the $V_{Th}$ is coming opposite in both. Why?