Question

A voltage source of 300 V has internal resistance of 4\(\Omega\) and supplies a load having the same resistance. The power absorbed by the load is

• A. 2510 W
• B. 1150 W
• C. 5625 W
• D. 5000 W

Hint:

For max power transfer, if load = source resistance, use \( I = \frac{V}{R_{total}} \), then calculate power with \( I^2R \).

Answer 1

The Correct Option is C

Step 1: Total resistance in the circuit \[ R_{total} = R_{internal} + R_{load} = 4\,\Omega + 4\,\Omega = 8\,\Omega \] Step 2: Calculate the current using Ohm’s law \[ I = \frac{V}{R_{total}} = \frac{300}{8} = 37.5\,A \] Step 3: Power absorbed by the load resistor \[ P_{load} = I^2 \cdot R_{load} = (37.5)^2 \cdot 4 = 1406.25 \cdot 4 = 5625\,W \] Hence, the power absorbed by the load is \( \boxed{5625\,W} \).