1. Circuit Analysis Strategy

The problem asks for the relationship between the voltage $V$ across the rheostat and the current $I$ flowing through it. The circuit consists of a network of batteries and resistors connected to the rheostat.

To solve this efficiently, we can treat the entire network connected to the rheostat terminals as a linear active two-terminal network. According to Thevenin’s Theorem, any such network can be replaced by a single equivalent voltage source ($E_{eq}$) in series with an equivalent resistance ($R_{eq}$).

The relationship between the terminal voltage $V$ and current $I$ for such a system is given by the linear equation: $$V = E_{eq} – I R_{eq}$$ Where $V$ is the potential difference across the load (rheostat) and $I$ is the current flowing from the network into the load.

2. Calculating Thevenin Equivalent of the Left Side (Terminal A)

The left side of the circuit connects to terminal A. It consists of two parallel branches connected to the ground (0V).

• Branch 1: A $100\text{ V}$ battery in series with $4\,\Omega$.
• Branch 2: A resistor of $4\,\Omega$ connected to ground (0V source).

The potential $V_A$ (open circuit voltage at the left terminal) is the average of the potentials of the sources weighted by their conductances (Millman’s Theorem):

$$ V_A = \frac{\frac{V_1}{R_1} + \frac{V_2}{R_2}}{\frac{1}{R_1} + \frac{1}{R_2}} $$ $$ V_A = \frac{\frac{100}{4} + \frac{0}{4}}{\frac{1}{4} + \frac{1}{4}} = \frac{25 + 0}{0.5} = 50\text{ V} $$

The equivalent resistance of the left side, $R_{th,L}$, is the two $4\,\Omega$ resistors in parallel:

$$ R_{th,L} = \frac{4 \times 4}{4 + 4} = 2\,\Omega $$

3. Calculating Thevenin Equivalent of the Right Side (Terminal B)

The right side connects to terminal B and consists of two identical parallel branches.

• Branch 3: A $100\text{ V}$ battery in series with $4\,\Omega$.
• Branch 4: A $100\text{ V}$ battery in series with $4\,\Omega$.

Using Millman’s Theorem for $V_B$:

$$ V_B = \frac{\frac{100}{4} + \frac{100}{4}}{\frac{1}{4} + \frac{1}{4}} = \frac{25 + 25}{0.5} = 100\text{ V} $$

The equivalent resistance of the right side, $R_{th,R}$, is the two $4\,\Omega$ resistors in parallel:

$$ R_{th,R} = \frac{4 \times 4}{4 + 4} = 2\,\Omega $$

4. Establishing the Relationship

The rheostat is connected between terminal A ($50\text{ V}, 2\,\Omega$) and terminal B ($100\text{ V}, 2\,\Omega$).

Total Equivalent Voltage ($E_{eq}$): Since the potential at B ($100\text{ V}$) is higher than A ($50\text{ V}$), the effective driving voltage is: $$ E_{eq} = V_B – V_A = 100 – 50 = 50\text{ V} $$

Total Equivalent Resistance ($R_{eq}$): The internal resistances of the left and right sides are in series with respect to the loop formed by the rheostat: $$ R_{eq} = R_{th,L} + R_{th,R} = 2 + 2 = 4\,\Omega $$

The terminal voltage $V$ across the rheostat is the open-circuit voltage minus the voltage drop due to its own internal resistance: $$ V = E_{eq} – I R_{eq} $$