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:

The correct answer is 10 volt.

Step-by-step Explanation:

To find the potential difference $V_A-V_B$ between points $A$ and $B$ in the given circuit, we can apply Kirchhoff's Voltage Law (KVL) starting from point $A$ and moving to point $B$.

From the provided image, we identify the following components and their values along the path from $A$ to $B$:

1. An inductor with self-inductance:
$L=1\text{H}$

2. A DC source (battery) with electromotive force (emf):
$E=5\text{V}$
Note that going from left to right, we move from the longer line (positive terminal) to the shorter, thicker line (negative terminal), which represents a drop in potential.

3. A resistor with resistance:
$R=2\text{Ω}$

The current $i$ flows from $A$ to $B$ with:
$i=2\text{A}$
The current is increasing at a rate of:
$\frac{di}{dt}=1\text{A/s}$

Now, let's write the potential equation as we move from $A$ to $B$:

- Starting potential is $V_A$.

- Since current is flowing from $A$ to $B$ and is increasing, the self-induced emf in the inductor opposes the increase in current. This results in a potential drop across the inductor of:
$-L\frac{di}{dt}$

- As we cross the battery from the positive to the negative terminal, there is a potential drop of:
$-E$

- As we cross the resistor in the direction of the current, there is a potential drop of:
$-iR$

Combining these potentials to reach the potential $V_B$ at point $B$:

$V_A-L\frac{di}{dt}-E-iR=V_B$

Rearranging the equation to solve for the potential difference $V_A-V_B$:

$V_A-V_B=L\frac{di}{dt}+E+iR$

Substituting the given values into the equation:

$V_A-V_B=\left(1\right)\left(1\right)+5+\left(2\right)\left(2\right)$

$V_A-V_B=1+5+4$

$V_A-V_B=10\text{V}$

Thus, the potential difference between point A and point B is 10 volt.