Question

In the circuit shown below, $V_s$ is constant voltage source and $I_L$ is a current load.

The value of $I_L$ that maximixes the power absorbed by the constant current load is

• A. $\frac{V_s}{4R}$
• B. $\frac{V_s}{2R}$
• C. $\frac{V_s}{4R}$
• D. $\infty$

Answer 1

The correct option is B $\frac{V_s}{2R}$

In maximum power transformation, half of the voltage drops across source resistance, remaining half across the load.

$\therefore$ Voltage across source (R)

$I_{L}R=\frac{V_s}{2}$

$I_L=\frac{V_s}{2R}$