# 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? – Andy aka Feb 25 '18 at 11:02
• Current ( $I_1$ ) is flowing from A to B – Suresh Feb 25 '18 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 $R_{TH}$ – Suresh Feb 25 '18 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.