Animated Solution for Physics - Electromagnetic Induction: The network shown in figure is part of a complete circuit. If at a certain instant the current ($I$) is 5A and is decreasing at a rate of $10^3$ A/s, then $V_B-V_A=....$ V

Imagine you are an electron, standing at point A, ready to embark on a journey through this circuit branch. Your goal is to reach point B, but the path is filled with obstacles and energy sources. To find out how much your potential energy changes by the time you reach B, we need to carefully track every single gain and loss along the way. This is the essence of Kirchhoff's Voltage Law (KVL).

Let's start our journey from point A, where our initial potential is $V_A$. We are moving in the direction of a $5\text{ A}$ current.

The first component we encounter is a $1\Omega$ resistor. Think of a resistor as a bumpy, friction-filled road. As you push through it in the direction of the current, you lose energy. This potential drop is calculated using Ohm's Law:

Next, we arrive at a $15\text{ V}$ battery. Notice the orientation of the battery symbol: we are entering the short, thick line (the negative terminal) and exiting the long, thin line (the positive terminal). This is like stepping onto an escalator that lifts you up to a higher energy state. Therefore, we gain potential:

Finally, we reach the most fascinating component: the $5\text{ mH}$ inductor. An inductor is like a magical water wheel that possesses inertia; it hates changes in flow. The problem states that the current is decreasing at a rate of $10^3\text{ A/s}$. Because the current is dropping, the inductor fights to keep it going, acting as a temporary battery pushing current forward.

Mathematically, the potential difference across an inductor is given by:

Here is where many students fall into a trap. You must respect the sign of the rate of change! Since the current is decreasing, $\frac{di}{dt}$ is negative ($-10^3\text{ A/s}$). Let's plug in the values:

The two negative signs cancel out, resulting in a positive potential change. The inductor is actively supplying energy to oppose the dying current!

Now, we have all the pieces of the puzzle. We simply string them together using our KVL equation from point A to point B:

Substituting the values we calculated:

Notice how the $5\text{ V}$ drop from the resistor is perfectly canceled out by the $5\text{ V}$ boost from the inductor. We are left with:

Rearranging this to find the potential difference $V_B-V_A$, we get our final, elegant answer:

By carefully respecting sign conventions and understanding the physical behavior of inductors, what seemed like a tricky circuit problem unravels into a beautiful sequence of logic.