Networks, Signals and Systems · Circuit Analysis · 1 mark · Multiple choice
In the circuit, $V_S$ is a constant voltage source and $I_L$ is a constant current load. The value of $I_L$ that maximizes the power absorbed by the load is
The resistor drops $I_LR$, so the load voltage is $V_S-I_LR$. Its absorbed power is $P=I_L(V_S-I_LR)$.
Differentiate with respect to $I_L$: $dP/dI_L=V_S-2RI_L$. The stationary point is $I_L=V_S/(2R)$. Because the second derivative is −2R<0, it is the maximum. Hence **B**.
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