Question

Output remains half of input when output resistance is

• A. \(\frac{R}{4}\)
• B. \(\frac{R}{6}\)
• C. \(\frac{R}{2}\)
• D. \(\frac{R}{8}\)

Hint:

Maximum power transfer occurs when load equals source resistance.

Answer 1

The Correct Option is C

Step 1: Model the source as a Thevenin equivalent.
Any practical source can be modeled as an ideal voltage $V$ in series with an internal resistance $R$. When a load $R_L$ is connected, the output voltage is $V_o = \dfrac{R_L}{R + R_L} \cdot V$.

Step 2: Apply the half-voltage condition.
For output to be half the input: $\dfrac{R_L}{R + R_L} = \dfrac{1}{2}$. Solving: $2R_L = R + R_L \Rightarrow R_L = R$. So the load equals the source internal resistance.

Step 3: Find the output resistance of the network.
With $R_L = R$ connected, the output resistance seen by any further load is $R \| R = R/2$. \[ \boxed{\dfrac{R}{2}} \]