Animated Solution for Physics - Electromagnetic Induction: A part of a complete circuit is shown in the figure. At some instant, the value of current $I$ is $1\text{ A}$ and it is decreasing at the rate of $10^2{\text{ As}}^{-1}$. The value of the potential difference $V_P-V_Q$ (in volt) at that instant, is ...........

Imagine you are walking along a path from point $P$ to point $Q$ in a circuit. Along this path, you encounter three distinct components: an inductor ($L=50\text{ mH}$), a battery ($30\text{ V}$), and a resistor ($R=2\Omega$).

Before we start our mathematical journey, we must establish the direction of the current. The problem states that the current $I$ is $1\text{ A}$ and is decreasing at a rate of $10^2\text{ A/s}$. Based on the visual cues in the diagram, the current is flowing from right to left, meaning it travels from $Q$ to $P$. This direction is crucial because it dictates the sign conventions we will use when applying Kirchhoff's Voltage Law (KVL).

Kirchhoff's Voltage Law

Let's apply KVL by traversing the circuit from left to right (from $P$ to $Q$). Since we are moving against the flow of the current, we must be very careful with our signs.

Starting at $P$, our potential is $V_P$. The first component we cross is the inductor. Because we are moving against the current, the potential change across the inductor is positive, given by $+L\frac{dI}{dt}$.

Next, we cross the battery. We enter the long line (positive terminal) and exit the short thick line (negative terminal). This represents a drop in potential, so we write $-30\text{ V}$.

Finally, we cross the resistor. Again, we are moving against the current, which means we are moving from a lower potential to a higher potential across the resistor. This gives us a potential gain of $+IR$.

Equating this entire journey to the potential at our destination, $Q$, we get our master equation:

We need to find the potential difference $V_P-V_Q$. Let's rearrange our master equation to isolate this term:

Now, it's time to substitute the raw values. We know $I=1\text{ A}$, $R=2\Omega$, and $L=50\text{ mH}=50\times 10^{-3}\text{ H}$. The most critical part is the rate of change of current. Since the current is decreasing, $\frac{dI}{dt}$ is negative, specifically $-100\text{ A/s}$.

Substituting these into our equation:

Let's compute the terms step-by-step. The resistor term is simply $2$. For the inductor term, multiplying $50\times 10^{-3}$ by $-100$ yields $-5$.

Subtracting a negative is the same as adding a positive.

And there we have it! The potential difference between points $P$ and $Q$ is exactly $33\text{ V}$.