How would the reading of the voltmeter $(V)$ change if it is connected between points $B$ and $C$? Justify your answer.

(N/A) The total resistance of the circuit is $R = R_1 + R_2 + R_3 = 1 \Omega + 3 \Omega + 2 \Omega = 6 \Omega$.
The total current in the circuit is $I = \frac{V}{R} = \frac{3 \text{ V}}{6 \Omega} = 0.5 \text{ A}$.
When the voltmeter is connected between $B$ and $C$,it measures the potential difference across the $3 \Omega$ resistor.
The potential difference across the $3 \Omega$ resistor is $V_{BC} = I \times R_{BC} = 0.5 \text{ A} \times 3 \Omega = 1.5 \text{ V}$.
Initially,the voltmeter was connected across the $1 \Omega$ resistor,measuring $V_{AB} = I \times R_{AB} = 0.5 \text{ A} \times 1 \Omega = 0.5 \text{ V}$.
Therefore,the reading of the voltmeter will increase from $0.5 \text{ V}$ to $1.5 \text{ V}$.