Consider the circuit given here. The potential difference $V_{B C}$ between the points $B$ and $C$ is

• A

  $1 \mathrm{V}$
• Correct answer
  B

  $0.5 \mathrm{V}$
• C

  $0 \mathrm{V}$
• D

  $-1 \mathrm{V}$

Solution

From circuit the current, $$ I=\frac{E}{R} $$ The equivalent resistance of above circuit $$ \begin{array}{l} =3+2+1 \\ =6 \mathrm{k} \Omega \\ =6 \times 10^{3} \Omega \\ E=3 \text { volt } \end{array} $$ $$ \begin{aligned} I &=\frac{3}{6 \times 10^{3}} \\ &=0.5 \times 10^{-3} \mathrm{A} \end{aligned} $$ Porential, $\begin{aligned} V_{A D} &=i R=0.5 \times 10^{-3} \times 3 \times 10^{3} \\ &=1.5 \mathrm{V} \end{aligned}$ Charge, $$ \begin{array}{l} \frac{1}{C}=\frac{1}{1}+\frac{1}{2}=\frac{3}{2} \\ C=\frac{2}{3} \\ Q=C \cdot V_{A D} \\ =\frac{2}{3} \times 1 \cdot 5 \\ =1 \mu \mathrm{C} \end{array} $$ Applying KVL (Kirchhoffs Voltage Law) from $B$ to $C$ $$ \begin{array}{c} v_{s}-0 \cdot 5 \times 10^{-3} \times 2 \times 10^{3}+\frac{1}{2}=V_{c} \\ V_{n}-V_{c}=1-\frac{1}{2}=0.5 \mathrm{V} \end{array} $$

Hence, the correct answer is option: 'B'.