Let the three identical resistances be $R$. Let the internal resistance of the voltmeter be $R_V$. The battery is ideal, so its internal resistance is zero. Let the EMF of the battery be $\mathcal{E}$.

Terminals A and D represent the entire external circuit connected to the battery. Since the battery is ideal, the voltage across A and D is simply the EMF of the battery, regardless of the voltmeter’s resistance (as the voltmeter is in parallel with the entire series combination, but measuring directly across the ideal source).

Reading $V_1 = 3.00 \, \text{V}$. Therefore:

$$ \mathcal{E} = V_1 = 3.00 \, \text{V} $$

The voltmeter is connected in parallel with the first resistor $R$ (between A and B). This combination is in series with the remaining two resistors ($R$ between B and C, and $R$ between C and D). The total resistance of the circuit becomes:

$$ R_{eq} = \left( \frac{R R_V}{R + R_V} \right) + R + R = \frac{R R_V}{R + R_V} + 2R $$

The voltage $V_2$ measured by the voltmeter is the potential difference across the parallel combination of $R$ and $R_V$. Using the voltage divider rule:

$$ V_2 = \mathcal{E} \cdot \frac{\frac{R R_V}{R + R_V}}{\frac{R R_V}{R + R_V} + 2R} $$

Substituting $\mathcal{E} = V_1$ and simplifying:

$$ V_2 = V_1 \cdot \frac{R R_V}{R R_V + 2R(R + R_V)} = V_1 \cdot \frac{R_V}{R_V + 2(R + R_V)} $$ $$ V_2 = V_1 \cdot \frac{R_V}{3R_V + 2R} $$

Rearranging to find the ratio $x = \frac{R_V}{R}$:

$$ \frac{V_1}{V_2} = \frac{3R_V + 2R}{R_V} = 3 + \frac{2R}{R_V} = 3 + \frac{2}{x} $$ $$ \frac{2}{x} = \frac{V_1}{V_2} – 3 = \frac{V_1 – 3V_2}{V_2} $$ $$ x = \frac{R_V}{R} = \frac{2V_2}{V_1 – 3V_2} \quad \dots(1) $$

Short-circuiting A and C means connecting them with a zero-resistance wire. This bypasses the resistors between A and B, and B and C for the main current path. The potential at A becomes equal to the potential at C ($V_A = V_C$).

The main circuit effectively consists only of the resistor between C and D connected to the battery. However, the voltmeter is connected between B and D.

• Node Potentials: Since $V_A = V_C$, and the resistors $R_{AB}$ and $R_{BC}$ form a path between A and C, the node B sits exactly in the middle of this shorted section. By symmetry, $V_B = V_A = V_C$.
• The Measuring Circuit: The voltmeter is connected between B and D. Effectively, it is measuring the potential difference between the node pair (A,C) and node D.

The circuit simplifies to the voltmeter (resistance $R_V$) being in parallel with the resistor group between nodes AC and B? No, let’s analyze the effective resistance.

With A and C shorted, the resistors $R_{AB}$ and $R_{BC}$ are in parallel between the node AC and node B. Their equivalent resistance is $R/2$. This combination is in series with the voltmeter. This entire branch (Resistance $R/2 + R_V$) is connected across the source terminals (since A is the positive terminal and D is the negative terminal).

Wait, the main source is connected to A and D. The resistor $R_{CD}$ is connected between C and D. Since A and C are shorted, $R_{CD}$ is directly across the source.

The voltmeter branch connects B to D. The path from A(source) to B goes through $R_{AB}$ (or $R_{BC}$ from C). Thus, we have a voltage divider consisting of the resistance from A/C to B (which is $R/2$) and the voltmeter resistance $R_V$.

Let the reading be $V_3$. This is the voltage across the voltmeter $R_V$.

$$ V_3 = \mathcal{E} \cdot \frac{R_V}{R_V + \frac{R}{2}} = V_1 \cdot \frac{R_V}{R_V + 0.5R} $$

Divide numerator and denominator by $R$ and substitute $x = R_V/R$:

$$ V_3 = V_1 \cdot \frac{x}{x + 0.5} $$

Substitute $x$ from equation (1):

$$ V_3 = V_1 \cdot \frac{\frac{2V_2}{V_1 – 3V_2}}{\frac{2V_2}{V_1 – 3V_2} + \frac{1}{2}} $$ $$ V_3 = V_1 \cdot \frac{4V_2}{4V_2 + (V_1 – 3V_2)} $$ $$ V_3 = V_1 \cdot \frac{4V_2}{V_1 + V_2} = \frac{4 V_1 V_2}{V_1 + V_2} $$