AB is a part of an electrical circuit (see figure). The potential difference " $\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$ ", at the instant when current $\mathrm{i}=2 \mathrm{~A}$ and is increasing at a rate of $1 \mathrm{amp} / \mathrm{second}$ is :

• A

  $5 \text { volt }$
• B

  $6 \text { volt }$
• C

  $9 \text { volt }$
• Correct answer
  D

  $10 \text { volt }$

Solution

$\begin{aligned} &\text { Apply K.V.L. }\\ &\begin{aligned} & \mathrm{V}_{\mathrm{A}}-\frac{\mathrm{LdI}}{\mathrm{dt}}-5-\mathrm{IR}=\mathrm{V}_{\mathrm{B}} \\ & \mathrm{~V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=\frac{\mathrm{LdI}}{\mathrm{dt}}+5+\mathrm{IR} \\ & =1 \times 1+5+2 \times 2 \\ & =1+5+4 \\ & =10 \mathrm{~V} \end{aligned} \end{aligned}$

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