A 24 V source with internal resistance 6 $\Omega$ feeds a load. Maximum power to load is:
Show Hint
Under the condition of maximum power transfer, exactly half of the total generated power is dissipated as heat within the source's internal resistance.
This means the maximum efficiency of power transfer is strictly limited to 50%.
Use the direct formula \(P_{max} = \frac{V_{th}^2}{4 R_{th}}\) to save time in competitive exams.
Step 1: Understanding the Question:
This question concerns the Maximum Power Transfer Theorem in DC networks.
The objective is to determine the maximum electrical power that can be delivered to a load resistor from a practical voltage source with a known internal resistance. Step 2: Key Formula or Approach:
According to the Maximum Power Transfer Theorem, the maximum power is transferred from a source to a load when the load resistance (\(R_L\)) is equal to the internal resistance (\(R_{in}\)) of the source:
\[ R_L = R_{in} \]
The maximum power (\(P_{max}\)) delivered to the load under this matched condition is calculated using:
\[ P_{max} = \frac{V_{in}^2}{4 R_{in}} \]
Step 3: Detailed Explanation:
• Identify the given parameters from the circuit diagram:
- Source voltage, \(V_{in} = 24\text{ V}\).
- Source internal resistance, \(R_{in} = 6\ \Omega\).
• Apply the condition for maximum power transfer to determine the load resistance:
\[ R_L = R_{in} = 6\ \Omega \]
• Calculate the total equivalent resistance of the series circuit:
\[ R_{total} = R_{in} + R_L = 6 + 6 = 12\ \Omega \]
• Find the current (\(I\)) flowing through the circuit under this condition:
\[ I = \frac{V_{in}}{R_{total}} = \frac{24}{12} = 2\text{ A} \]
• Calculate the power dissipated in the load resistor \(R_L\):
\[ P_{max} = I^2 \cdot R_L = (2)^2 \times 6 = 4 \times 6 = 24\text{ W} \]
• Alternatively, using the direct formula:
\[ P_{max} = \frac{V_{in}^2}{4 R_{in}} = \frac{24^2}{4 \times 6} = \frac{576}{24} = 24\text{ W} \]
Step 4: Final Answer:
The maximum power transferred to the load is 24 W.
