In the network of figure,the maximum power is delivered to \$R_L\$ if its value is ______ \$ ?\$
simulate this circuit – Schematic created using CircuitLab
where, \$ I_1 \$ is the current flowing through \$ AB \$ via \$ A \$ to \$ B \$
My Approach:
we know for maximum power delivered to \$ R_L \$ :
\$ R_L = R_{TH}\$ where \$ R_{TH} \$ is the Thevenin's Resistance
To calculate \$ R_{TH} \$ :
Process 1:
\$ R_{TH}= \frac{V_{TH}}{I_N} \$ or \$ R_{TH}= \frac{V_{OC}}{I_{SC}} \$
For \$ V_{OC} \$ :
Applying KCL at node \$A\$ : $$0.5 \times I_1 = \frac{V_{OC}}{20} + I_1$$ $$\implies V_{OC}=- 10I_1$$ and $$I_1=\frac{V_{OC}-50}{40}$$ so,after solving for \$ V_{OC} \$ we get: $$V_{OC}=10V$$
In a similar fashion we calculate \$ I_{SC} \$ by shorting \$ R_L \$ and get $$I_{SC}=0.625A$$ $$\therefore R_{TH}= \frac{10}{0.625}=16\Omega$$
Hence,maximum power is delivered to \$R_L\$ if its value is \$ 16\Omega \$
Process2:
Here, \$ R_{TH}=\frac{V_a}{I_a} \$ where \$ I_a \$ is the current flowing out from \$ V_a \$ voltage source
so, \$ I_1 = \frac{100-50}{40}= \frac{5}{4} \Omega \$
so applying KCL at node \$ A \$ : $$0.5 \times \frac{5}{4} + I_a = \frac{100}{20} + \frac{5}{4}$$ $$\implies I_a=\frac{45}{8} A$$ $$\therefore R_{TH}=\frac{100}{\frac{45}{8}}= 17.78\Omega$$
Hence,maximum power is delivered to \$R_L\$ if its value is \$ 17.78\Omega \$
Thus,which process is right? please help....