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

Correct answer: D. 10 volt aligned & Apply K.V.L. & aligned & V_ A - LdI dt -5-IR=V_ B & V_ A -V_ B = LdI dt +5+IR & =1 × 1+5+2 × 2 & =1+5+4 & =10 V aligned aligned

NEET 2025

Exam: NEET
Year: 2025
Subject: Physics
Chapter: Electromagnetic Induction

NEET 2025 · Physics · Electromagnetic Induction

AB is a part of an electrical circuit (see figure). The potential...

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 :

AB is a part of an electrical circuit (see figure). The potential difference " V_ A -V_ B ", at the instant when

Options

A $5$ $\text{ volt }$
B $6$ $\text{ volt }$
C $9$ $\text{ volt }$
D $10$ $\text{ volt }$

Correct answer

D. $10$ $\text{ volt }$

Step-by-step 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}$

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