Step 1: Thevenin Equivalent Circuit

To find the maximum power developed in the variable resistance $R$, we first determine the Thevenin equivalent circuit of the network as seen from the terminals of $R$.

The network consists of two voltage sources in parallel branches. We apply Millman’s Theorem (or parallel battery formula) to find the Thevenin voltage ($V_{th}$) and Thevenin resistance ($R_{th}$).

Thevenin Resistance ($R_{th}$):
This is the resistance looking into the terminals with voltage sources shorted (replaced by wires). $R_1$ becomes parallel to $R_2$.

$$ R_{th} = R_1 \parallel R_2 = \frac{R_1 R_2}{R_1 + R_2} $$

Thevenin Voltage ($V_{th}$):
Using Millman’s theorem:

$$ V_{th} = \frac{\frac{V_1}{R_1} + \frac{V_2}{R_2}}{\frac{1}{R_1} + \frac{1}{R_2}} = \frac{\frac{V_1 R_2 + V_2 R_1}{R_1 R_2}}{\frac{R_1 + R_2}{R_1 R_2}} = \frac{V_1 R_2 + V_2 R_1}{R_1 + R_2} $$

Step 2: Maximum Power Transfer Theorem

According to the Maximum Power Transfer Theorem, maximum power is dissipated in the load resistance $R$ when it is equal to the internal (Thevenin) resistance of the source.

$$ R = R_{th} = \frac{R_1 R_2}{R_1 + R_2} $$

Step 3: Calculating Maximum Power

The maximum power is given by the formula:

$$ P_{\text{max}} = \frac{V_{th}^2}{4 R_{th}} $$

Substituting the values derived above:

$$ P_{\text{max}} = \frac{\left( \frac{V_1 R_2 + V_2 R_1}{R_1 + R_2} \right)^2}{4 \left( \frac{R_1 R_2}{R_1 + R_2} \right)} $$ $$ P_{\text{max}} = \frac{(V_1 R_2 + V_2 R_1)^2}{(R_1 + R_2)^2} \times \frac{R_1 + R_2}{4 R_1 R_2} $$ $$ P_{\text{max}} = \frac{(V_1 R_2 + V_2 R_1)^2}{4 R_1 R_2 (R_1 + R_2)} $$