An LCR circuit is equivalent to a damped pendulum. In an LCR circuit, the capacitor is charged to Q0

Q: Why is current $i = -dq/dt$ and not $+dq/dt$?

Because the capacitor is acting as the source of energy and is discharging. The charge $q$ on the plates is decreasing with time, making $dq/dt$ negative. To keep the physical current $i$ positive, we must define it as $i = -dq/dt$.

Q: How do we know the solution is an exponential decay?

The equation $L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = 0$ is a standard second-order linear homogeneous differential equation. For a lightly damped system (small R), the solution is a sinusoidal wave multiplied by an exponential decay envelope $e^{-Rt/2L}$.

Q: What does 'electrical inertia' mean?

In mechanics, mass (inertia) opposes changes in velocity. In electricity, an inductor opposes changes in current (due to Lenz's Law). Therefore, a larger inductor makes it harder for the circuit to lose its oscillating current quickly.

Q: Why does a larger $L$ mean slower decay?

Look at the decay term $e^{-Rt/L}$. The inductance $L$ is in the denominator of the exponent. A larger $L$ makes the fraction $R/L$ smaller, meaning the value of the exponent approaches zero, and $e^0 = 1$. Thus, the charge decays less rapidly.

Q: What would happen if the resistance $R$ was zero?

If $R=0$, the decay term $e^{-0}$ becomes exactly 1. The circuit would be an ideal, undamped LC oscillator, and the charge would oscillate back and forth forever without any loss of maximum amplitude.

Question

An LCR circuit is equivalent to a damped pendulum. In an LCR circuit, the capacitor is charged to $Q_0$ and then connected to the $L$ and $R$ as shown below. If a student plots graphs of the square of maximum charge ($Q_{	ext{Max}}^2$) on the capacitor with time ($t$) for two different values $L_1$ and $L_2$ ($L_1 > L_2$) of $L$, then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)

Sigma Percentile · Accepted Answer

### The Electrical Pendulum

Imagine a simple pendulum swinging back and forth. If you submerge it in water, the swings get smaller and smaller until it eventually stops. This is a classic damped mechanical oscillator.

Fascinatingly, an LCR circuit behaves in the exact same mathematical way! When a fully charged capacitor is connected to an inductor and a resistor, the energy sloshes back and forth between the electric field of the capacitor and the magnetic field of the inductor. However, the resistor acts like the "water" in our pendulum analogy—it dissipates energy as heat, causing the oscillations to decay over time.

### The Master Equation

To understand exactly how the charge decays, we need to write down the governing equation of the circuit. By applying Kirchhoff's Voltage Law (KVL) around the loop, we sum the potential drops across the capacitor, resistor, and inductor to zero:

$$\frac{q}{C} - iR - L\frac{di}{dt} = 0$$

Because the capacitor is discharging, the current $i$ is the negative rate of change of charge, meaning $i = -\frac{dq}{dt}$. Substituting this into our KVL equation gives us:

$$\frac{q}{C} - \left(-\frac{dq}{dt}ight)R - L\frac{d}{dt}\left(-\frac{dq}{dt}ight) = 0$$

Cleaning up the negative signs and rearranging the terms, we arrive at a beautiful second-order linear differential equation:

$$L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{q}{C} = 0$$

Dividing by $L$, we get the standard form:

$$\frac{d^2q}{dt^2} + \frac{R}{L}\frac{dq}{dt} + \frac{q}{LC} = 0$$

### The Exponential Decay

This differential equation is mathematically identical to that of a damped mechanical oscillator. The solution for the amplitude (the maximum charge $Q_{	ext{max}}$ during each cycle) decays exponentially over time. The formula for this decaying amplitude is:

$$Q_{	ext{max}}(t) = Q_0 e^{-\frac{Rt}{2L}}$$

However, the question specifically asks us to analyze the **square** of the maximum charge, $Q_{	ext{max}}^2$. Let's square our amplitude equation. When you square an exponential, you simply multiply the exponent by 2:

$$Q_{	ext{max}}^2(t) = Q_0^2 e^{-\frac{Rt}{L}}$$

### Analyzing the Graphs

Now comes the crucial physics intuition. We are given two different inductances, $L_1$ and $L_2$, with the condition that **$L_1 > L_2$**.

Look closely at the exponent: $-\frac{Rt}{L}$. The inductance $L$ is in the denominator. A larger value of $L$ makes the entire fraction $\frac{R}{L}$ smaller. This means the negative exponent is smaller in magnitude, resulting in a **slower exponential decay**.

Think of inductance as **electrical inertia**. Just as a massive object is harder to slow down, a circuit with a large inductance strongly opposes changes in current, causing the energy (and thus the charge) to persist for a longer time.

Because $L_1 > L_2$, the charge for $L_1$ will decay slower than the charge for $L_2$. If we pick any arbitrary time $t_0$ on the graph, the curve for $L_1$ must be physically higher than the curve for $L_2$. Mathematically, $Q_1^2 > Q_2^2$.

Looking at the given options, only the first graph correctly shows the curve for $L_1$ decaying slower and remaining above the curve for $L_2$.

Select Answer:

Visualized Solution

Visual Anchor: The LCR Circuit

Logic Bridge: Kirchhoff's Voltage Law

Raw Setup: Current and Charge Relation

Atomic Compute: The Differential Equation

Atomic Compute: Amplitude Decay

Atomic Compute: Squaring the Amplitude

Atomic Compute: Analyzing Inductance

Final Answer: Graph Selection

The Way Forward

The Sigma Insight: Free, Forced, and Damped Oscillations

The Electrical Pendulum

The Master Equation

The Exponential Decay

Analyzing the Graphs

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