Electrical problem;maximum power transfer;confusion

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 $17.78\Omega$

• Which direction does the current flow? Mark it on your diagrams. Process 2 doesn’t have the same circuit so how might it be the same? Commented Feb 25, 2018 at 11:02
• Current ( $I_1$ ) is flowing from A to B Commented Feb 25, 2018 at 11:11
• @ Andy aka In process 2 : we have added an arbitrary ( $V_a$ ) Voltage source in place of $R_{L}$ to find $R_{TH}$ ........i guess it's another process to evaluate $R_{TH}$ Commented Feb 25, 2018 at 11:14

I know this answer is a bit old but I saw it and wanted to answer it in case the OP is still waiting for an answer.

Checking Power for Both Processes

My first thought when seeing this question was to calculate the power at 16Ω at 17.78Ω and see which is larger. This way I know which process to focus in on for looking for the error.

Using the original circuit, here is the equation using KCL at Node A:

$$-0.5I_1 + \frac{V_a}{20} + \frac{V_a}{R_L} + I_1 = 0$$

Note: I usually write KCL such that all of the currents are coming out of the node. This is why there is a negative for the dependent source.

There are two unknowns in the equation above so we define $I_1$ in terms of $V_a$:

$$I_1 = \frac{V_a - 50}{40}$$

Plugging $I_1$ into the KCL equation and solving for $V_a$ gives

$$V_a = \frac{0.625}{0.0625 + \frac{1}{R_L}} \\ \\ V_a(R_L = 16 \Omega) = 5V \\ V_a(R_L = 17.78 \Omega) = 5.2634V$$

The power can easily be calculated with the voltage and load resistance known ($P = \frac{V^2}{R}$):

$$P(R_L = 16 \Omega) = 1.5625V \\ P(R_L = 17.78 \Omega) = 1.558V$$

This leads me to believe that the first process you used is correct and there is an error in your second process.

What's the Issue Then?

It's been awhile since I've had circuits 101 and to be honest, I haven't used thevenin resistances much since then.

According to AllAboutCircuits, when using a test source, you must turn off independent sources!

Let us repeat the calculation with the dependent source off.

simulate this circuit – Schematic created using CircuitLab

We can calculate $I_1$ to be $100 / 40 \Omega = 2.5A$.

Applying KCL at node A:

$$-I_1 * 0.5 + 100 / 20 - I_a + I_1 = 0 \\ -1.25A + 5A + 2.5A = I_a \\ I_a = 6.25A \\ R_{th} = \frac{V_a}{I_a} = \frac{100}{6.25} = 16 \Omega$$

Long story short, you forgot to turn off the independent source.