Three voltmeters, all having different resistances, are joined as shown. When some potential difference is applied across A and B, their readings are $V_{1}, V_{2}, V_{3}$

V_{1} = V_{2}

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V_{1} \neq V_{2}

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V_{1} + V_{2} = V_{3}

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V_{1} + V_{2} > V_{3}

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Solution

The correct option is C $V_{1} + V_{2} = V_{3}$
Here, three voltmeters are joined as shown in given figure.
We know,
$V = I R$
Also, all voltmeters have different resistances, then for same current $V \propto R$
Hence, $V_{1} \propto R_{1}$
similarly, $V_{2} \propto R_{2}$ and $V_{3} \propto R_{3}$
$\Rightarrow V_{1} \neq V_{2}$
When some potential difference is applied the current will divide into two branches i.e. say $I_{1}$ in one having $V_{1}$ and $V_{2}$ in series and say $I_{2}$ in other having $V_{3}$ parallel to both of them.
Now, applying KVL to the loop
$I_{1} R_{1} + I_{1} R_{2} - I_{2} V_{3} = 0$
$V_{1} + V_{2} - V_{3} = 0$
$V_{1} + V_{2} = V_{3}$

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Three voltmeters, all having different resistances, are joined as shown. When some potential difference is applied across A and B, their readings are V1,V2,V3

Three voltmeters, all having different resistances, are joined as shown. When some potential difference is applied across A and B, their readings are $V_{1}, V_{2}, V_{3}$