AB is a part of an electrical circuit (see figure). The potential difference “VA – VB”, at the instant when current i = 2 A and is increasing at a rate of 1 amp/second is:

To find the potential difference $V_A-V_B$ between points A and B, we can apply Kirchhoff's Voltage Law (KVL) along the branch from A to B.

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

1. An inductor with self-inductance $L=1\text{H}$.
2. A DC voltage source (battery) with potential difference $E=5\text{V}$, where the longer line represents the positive terminal (left) and the shorter line represents the negative terminal (right).
3. A resistor with resistance $R=2\text{Ω}$.
4. The current $i=2\text{A}$ flows from A to B, and it is increasing at a rate of $\frac{di}{dt}=1\text{A/s}$.

Let us write down the potential equation as we move from A to B along the direction of the current:

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

Substituting the given values into the equation:

$V_A-\left(1\text{H}\right)\left(1\text{A/s}\right)-5\text{V}-\left(2\text{A}\right)\left(2\text{Ω}\right)=V_B$

Simplifying the potential drops:

$V_A-1-5-4=V_B$

$V_A-10=V_B$

Rearranging the equation to find the potential difference $V_A-V_B$:

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

Therefore, the potential difference is 10 volt.