In the circuit shown here, the point C is kept connected to point A till the current flowing through

Q: Why can't we just use the formulas $V_R = I_0 R e^{-Rt/L}$ and $V_L = -I_0 R e^{-Rt/L}$?

You absolutely can! If you calculate them, you'll see that $V_R$ and $V_L$ are equal in magnitude but opposite in sign at every instant. Dividing them gives $-1$. However, using KVL directly saves you from doing any of that math!

Q: What is the significance of $t = L/R$ in this question?

It's a classic distractor! $L/R$ is the time constant of the circuit, which is the time it takes for the current to decay to about 37% of its initial value. The examiner put it there to trick you into doing unnecessary calculations.

Q: Why is $V_L$ negative when $V_R$ is positive?

During discharge, the inductor acts as a source of energy, trying to maintain the current. It pushes current through the resistor. By passive sign convention, if the resistor consumes energy (positive voltage drop), the inductor providing the energy will have a negative voltage drop in the loop equation.

Q: Does the initial current $I_0$ matter for this ratio?

Not at all. Whether the inductor was fully charged or only partially charged, the KVL equation $V_R + V_L = 0$ holds true for the closed loop, making the ratio independent of the initial current.

Q: What if the switch was kept at A? What would the ratio be?

If the switch is at A, the battery $V$ is in the loop. KVL becomes $V - V_R - V_L = 0$. The ratio $V_R/V_L$ would then depend on time, as $V_R$ increases and $V_L$ decreases as the inductor charges.

Question

In the circuit shown here, the point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterward, suddenly point C is disconnected from point A and connected to point B at time $t = 0$. Ratio of the voltage across resistance and the inductor at $t = L/R$ will be equal to

Options:

(A) $\frac{e}{1-e}$

(B) $1$

(C) $-1$

(D) $\frac{1-e}{e}$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two States

Imagine you are observing a classic electrical circuit. Initially, the switch is connected to point A. The battery is hard at work, pumping current through the resistor and the inductor.

The inductor, being the stubborn component it is, slowly builds up its magnetic field until the current reaches a steady, constant state.

But then, at exactly $t = 0$, the switch is violently thrown from point A to point B. The battery is suddenly cut off from the rest of the circuit. What remains is a simple, isolated closed loop containing only the resistor and the fully charged inductor.

### The Distractor: Time is an Illusion

The question asks a very specific and seemingly complex question: "What is the ratio of the voltage across the resistance to the voltage across the inductor at exactly $t = L/R$?"

If you are like most students, your brain immediately jumps to the formulas for $L-R$ discharging circuits. You might recall that the current decays exponentially according to $i(t) = I_0 e^{-Rt/L}$.

You might then plan to calculate the voltage across the resistor as $V_R = iR$ and the voltage across the inductor as $V_L = L \frac{di}{dt}$. Finally, you would plug in $t = L/R$ to find the ratio.

While this method is mathematically sound, it is a massive trap! The examiner has deliberately given you a specific time to lure you into a maze of unnecessary calculations.

### The Masterstroke: Kirchhoff's Voltage Law

Let's take a step back and look at the circuit through the lens of fundamental physics. Once the switch is connected to point B, we have a closed loop with no external power source.

This is where **Kirchhoff's Voltage Law (KVL)** comes to our rescue. KVL states that the algebraic sum of all potential differences in any closed loop must be exactly zero.

Applying this to our isolated $R-L$ loop, we can write the master equation:

$$V_R + V_L = 0$$

### The Elegant Conclusion

The beauty of this equation lies in its simplicity. If the sum of the two voltages is zero, then one must be the exact negative of the other:

$$V_R = -V_L$$

By simply dividing both sides by $V_L$, we find our answer:

$$\frac{V_R}{V_L} = -1$$

Notice what is missing from this final result? **Time!**

The ratio of the voltages is $-1$ at $t = 1$ second, at $t = 10$ seconds, and yes, even at $t = L/R$. The ratio is a fundamental constant of the discharging loop. By trusting in the core principles of physics, we bypassed the complex calculus entirely and arrived at the elegant truth in seconds.

Select Answer:

Visualized Solution

Analyzing the Circuit States

Kirchhoff's Voltage Law (KVL)

Applying KVL to the Discharging Loop

Calculating the Ratio

Independence of Time

The Distractor Trap

The Sigma Insight: Self and Mutual Inductance

The Setup

The Distractor

The Masterstroke

The Elegant Conclusion

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