Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time tak

Q: Why does energy decay twice as fast as charge?

Because energy $U$ is proportional to the square of the charge ($q^2$). When you square an exponential function like $e^{-t/\tau}$, the exponent gets multiplied by 2, resulting in $e^{-2t/\tau}$, which represents a decay rate that is twice as fast.

Q: Can I use $U = \frac{1}{2}CV^2$ instead of the charge formula?

Absolutely! The voltage across the capacitor also decays exponentially as $V(t) = V_0 e^{-t/\tau}$. Squaring the voltage will give you the exact same $e^{-2t/\tau}$ factor for the energy decay.

Q: What exactly is the time constant $\tau$?

The time constant $\tau = RC$ is a measure of how quickly the circuit responds. It is the time required for the charge (or voltage) to drop to $1/e$ (about 36.8%) of its initial value during discharge.

Q: Why do we use the natural log ($\ln$) to solve this?

Because the decay is modeled using the mathematical constant $e$ (Euler's number). The natural logarithm is the inverse operation of the exponential function base $e$, allowing us to easily solve for the time variable in the exponent.

Q: Does the final ratio depend on the specific values of the resistor $R$ and capacitor $C$?

No, it doesn't! As you can see in the final step, the time constant $\tau$ (which contains $R$ and $C$) completely cancels out when we take the ratio. The ratio is purely a mathematical consequence of the exponential decay laws.

Question

Let $C$ be the capacitance of a capacitor discharging through a resistor $R$. Suppose $t_1$ is the time taken for the energy stored in the capacitor to reduce to half its initial value and $t_2$ is the time taken for the charge to reduce to one-fourth its initial value. Then, the ratio $t_1/t_2$ will be

Options:

(A) 1

(B) $\frac{1}{2}$

(C) $\frac{1}{4}$

(D) 2

Sigma Percentile · Accepted Answer

### The Discharging Capacitor

Imagine a fully charged capacitor, brimming with electrical energy, suddenly connected across a resistor. The moment the circuit is completed, the charge starts to flow, and the capacitor begins to discharge. This is the classic RC circuit, a fundamental concept in physics that beautifully demonstrates exponential decay.

### The Tale of Two Decays

As the capacitor discharges, the charge on its plates doesn't just drop linearly; it decays exponentially. The charge $q$ at any time $t$ is governed by the equation:
$$q(t) = q_0 e^{-t/	au}$$
where $q_0$ is the initial charge and $	au = RC$ is the time constant of the circuit.

But what about the energy? The energy $U$ stored in a capacitor is given by $U = \frac{q^2}{2C}$. If we substitute our exponentially decaying charge into this formula, we get:
$$U(t) = \frac{(q_0 e^{-t/	au})^2}{2C} = \frac{q_0^2}{2C} e^{-2t/	au} = U_0 e^{-2t/	au}$$
Notice the crucial difference here! The energy also decays exponentially, but because of the squared term, there is a factor of $2$ in the exponent. This means the energy decays **twice as fast** as the charge.

### Calculating the Times

The problem presents us with two specific milestones in the capacitor's discharge journey. First, we are told that at time $t_1$, the energy reduces to exactly half of its initial value. Let's set up the equation for this:
$$\frac{U_0}{2} = U_0 e^{-2t_1/	au}$$
The initial energy $U_0$ cancels out. Taking the natural logarithm on both sides, we find:
$$\ln\left(\frac{1}{2}ight) = -\frac{2t_1}{	au} \implies -\ln 2 = -\frac{2t_1}{	au} \implies t_1 = \frac{	au}{2} \ln 2$$

Next, we are told that at time $t_2$, the charge on the capacitor reduces to one-fourth of its initial value. We set up a similar equation for the charge:
$$\frac{q_0}{4} = q_0 e^{-t_2/	au}$$
Solving for $t_2$, the initial charge $q_0$ cancels out. Taking the natural log, and remembering that $\ln(1/4) = -2\ln 2$, we get:
$$\ln\left(\frac{1}{4}ight) = -\frac{t_2}{	au} \implies -2\ln 2 = -\frac{t_2}{	au} \implies t_2 = 2	au \ln 2$$

### The Final Ratio

Finally, we need to find the ratio of $t_1$ to $t_2$. Dividing our expression for $t_1$ by our expression for $t_2$:
$$\frac{t_1}{t_2} = \frac{\frac{	au}{2} \ln 2}{2	au \ln 2}$$
The time constant $	au$ and the $\ln 2$ terms beautifully cancel out, leaving us with exactly:
$$\frac{t_1}{t_2} = \frac{1}{4}$$
This elegant result highlights the interconnected yet distinct rates at which charge and energy dissipate in an RC circuit.

Select Answer:

Visualized Solution

$R C$ Discharging Circuit

Charge Decay Formula

Energy Decay Formula

Setting up $t_{1}$

Solving for $t_{1}$

Setting up $t_{2}$

Solving for $t_{2}$

Final Ratio

The Way Forward

The Sigma Insight: RC Circuit

The Discharging Capacitor

The Tale of Two Decays

Calculating the Times

The Final Ratio

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