Question

The efficiency of a circuit at maximum power transfer condition is

• A. 100%
• B. 50%
• C. 25%
• D. 75%

Hint:

The Maximum Power Transfer Theorem is used to maximize power output in communication systems (such as impedance matching for antennas or speakers), rather than to maximize efficiency. In high-power electrical grid distribution networks, engineers avoid this condition because a 50% energy loss within the power plant is highly inefficient!

Answer 1

The Correct Option is B

Concept: The Maximum Power Transfer Theorem states that a linear, resistive DC network will deliver the maximum possible amount of electrical power to an external load when the resistance of that load ($R_L$) matches the internal Thevenin equivalent resistance ($R_{th}$) of the source network looking back from the terminals: \[ R_L = R_{th} \] Let us derive the electrical efficiency ($\eta$) under this maximum power transfer condition.

Step 1: Set up the total circuit current. Consider a simple loop consisting of a Thevenin source voltage $V_{th}$, an internal source resistance $R_{th}$, and a load resistance $R_L$ connected in series. The total loop current $I$ is given by Ohm's law: \[ I = \frac{V_{th}}{R_{th} + R_L} \]

Step 2: Express total input power and output load power.
• The total power generated and supplied by the source voltage ($P_{\text{in}}$) is: \[ P_{\text{in}} = V_{th} \cdot I \]
• The actual useful power delivered to and consumed by the load resistor ($P_{\text{out}}$ or $P_L$) is: \[ P_{\text{out}} = I^2 \cdot R_L \]

Step 3: Apply the maximum power condition ($R_L = R_{th}$). Substituting $R_L = R_{th}$ into the current expression gives: \[ I = \frac{V_{th}}{R_{th} + R_{th}} = \frac{V_{th}}{2R_{th}} \] Now, let us calculate the total input power under this condition: \[ P_{\text{in}} = V_{th} \cdot \left(\frac{V_{th}}{2R_{th}}\right) = \frac{V_{th}^2}{2R_{th}} \] Next, let us calculate the output power delivered to the load: \[ P_{\text{out}} = I^2 \cdot R_L = \left(\frac{V_{th}}{2R_{th}}\right)^2 \cdot R_{th} = \frac{V_{th}^2}{4R_{th}^2} \cdot R_{th} = \frac{V_{th}^2}{4R_{th}} \]

Step 4: Compute the efficiency ($\eta$). Efficiency is defined as the ratio of useful output power to total input power: \[ \eta = \frac{P_{\text{out}}}{P_{\text{in}}} = \frac{\frac{V_{th}^2}{4R_{th}}}{\frac{V_{th}^2}{2R_{th}}} = \frac{2R_{th}}{4R_{th}} = \frac{1}{2} \] Converting this fraction into a percentage: \[ \eta % = \frac{1}{2} \times 100% = 50% \] Thus, at the maximum power transfer condition, the circuit efficiency is exactly 50%. The remaining 50% of the energy is lost as heat inside the source's internal resistance ($R_{th}$). This matches Option (2).