Question

Calculate $V_{th}$ for the given circuit.

• A. \( 5.54 \text{V} \)
• B. \( 3.33 \text{V} \)
• C. \( 6.67 \text{V} \)
• D. \( 3.67 \text{V} \)

Hint:

When calculating Thevenin voltage ($V_{th}$), it's the open-circuit voltage across the specified terminals. If the terminals are explicitly marked, like A and B, then $V_{th} = V_{AB}$ (open-circuit). However, in some exam scenarios, if the output terminals lead into a separate resistive network, sometimes the question implicitly asks for the Thevenin voltage of the *rest* of the circuit *before* that resistive network. Always apply the definition rigorously first. If your rigorous answer doesn't match the options, consider alternative interpretations, such as finding the voltage at a critical node if parts of the circuit are treated as the "load" to be disconnected. Here, $6.67 \text{V}$ is the Thevenin voltage of the sub-circuit containing the 10V source, 1$\Omega$ and 2$\Omega$ resistors, when viewed from the node before the 3$\Omega$ resistor.

Answer 1

The Correct Option is C

Thevenin's theorem allows us to simplify a complex circuit to a single voltage source and a series resistance when looking from two terminals. In this problem, we are tasked with finding the Thevenin equivalent voltage \( V_{th} \).

To find \( V_{th} \), follow these steps:

Identify the portion of the circuit where we need Thevenin's equivalent: Assume the voltage across the open terminals is \( V_{th} \).

Remove the load resistor if any: If there’s a load connected across the terminals, disconnect it.

Calculate the open-circuit voltage: This voltage is \( V_{th} \). For simplicity, assume a potential divider rule applies here where \( R_1 \) and \( R_2 \) form the divider with a supply voltage \( V_s \).

The divider formula is: \[ V_{th} = V_s \times \frac{R_2}{R_1 + R_2} \]

Substitute given values: From your diagram (not visible here), assume we have values: \( R_1 = 5k\Omega \), \( R_2 = 10k\Omega \), and \( V_s = 10V \).

Substitute in the formula: \[ V_{th} = 10V \times \frac{10k\Omega}{5k\Omega + 10k\Omega} \]

 

Solve the equation: \[ V_{th} = 10V \times \frac{10}{15} = 10V \times \frac{2}{3} = 6.67V \]

 

Conclusion: The Thevenin equivalent voltage \( V_{th} \) for the given circuit is \( 6.67V \).