Question 319: Electromagnetic Induction

Question

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

Accepted Answer

$$\begin{aligned}
&	ext { Given, } l=2 \mathrm{~A} 	ext { and } \frac{d i}{d t}=+1 \mathrm{~A} / \mathrm{s}\
&\begin{aligned}
& V_A-L \frac{d i}{d t}-5-i 	imes 2=V_B \
& \Rightarrow V_A-1 	imes 1-5-2 	imes 2=V_B \
& \Rightarrow V_A-V_B=10 	ext { volt }
\end{aligned}
\end{aligned}$$

Answer

9 volt

Answer

10 volt

Answer

5 volt

Answer

6 volt

Electromagnetic Induction

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

Choose the correct answer: