Interpret the circuit

The standard arrangement for this question is that voltmeters $V_1$ and $V_2$ are connected across two parts of a series combination between terminals $A$ and $B$, while $V_3$ is connected across the whole combination.

Since voltmeters have finite internal resistances, let their internal resistances be: $R_1,\; R_2,\; R_3$ respectively.

Key idea

A voltmeter reads the potential difference across the points to which it is connected.

Suppose the resistor-equivalent/path between $A$ and the middle point is across which $V_1$ is connected, and between the middle point and $B$ is across which $V_2$ is connected. Then $V_3$ is connected across the full terminals $A$ and $B$.

Potential difference is a state function: the total potential difference across $A$ and $B$ is equal to the sum of the potential differences across the two successive parts.

Therefore, $V_{AB}=V_{AM}+V_{MB}$ where $V_1=V_{AM},\quad V_2=V_{MB},\quad V_3=V_{AB}$

Hence, $V_3=V_1+V_2$

Why internal resistances do not spoil this relation

Although the voltmeters have different internal resistances and may draw different currents, each voltmeter still measures the potential difference across its own terminals.

Since the points across which $V_1$ and $V_2$ are connected are consecutive segments spanning the same endpoints as $V_3$, Kirchhoff’s voltage law gives: $V_1+V_2=V_3$

Check options

• A: $V_1=V_2$
  Not necessarily true.
• B: $V_1\neq V_3-V_2$
  False, because from above: $V_1=V_3-V_2$
• C: $V_1+V_2=V_3$
  True.
• D: $V_1+V_2>V_3$
  False.

Final answer

The correct option is: $\boxed{C}$