1. Discharge equations for a capacitor

When a capacitor of capacitance $C$ discharges through a resistor $R$, the charge at time $t$ is

$q(t)=q_0 e^{-t/RC}$

where $q_0$ is the initial charge.

The energy stored in the capacitor is

U=rac{q^2}{2C}

So during discharge,

$U(t)=U_0 e^{-2t/RC}$

because energy depends on the square of charge.

2. Find $t_1$ from energy becoming half

Given that energy becomes half of its initial value:

$U(t_1)=\frac{U_0}{2}$

Using

$U(t)=U_0 e^{-2t/RC}$

we get

$e^{-2t_1/RC}=\frac{1}{2}$

Taking natural log:

$-\frac{2t_1}{RC}=\ln\left(\frac{1}{2}\right)=-\ln 2$

Hence,

$t_1=\frac{RC}{2}\ln 2$

3. Find $t_2$ from charge becoming one-fourth

Given that charge becomes one-fourth of initial value:

$q(t_2)=\frac{q_0}{4}$

Using

$q(t)=q_0 e^{-t/RC}$

we get

$e^{-t_2/RC}=\frac{1}{4}$

Taking natural log:

$-\frac{t_2}{RC}=\ln\left(\frac{1}{4}\right)=-\ln 4=-2\ln 2$

Hence,

$t_2=2RC\ln 2$

4. Compute the ratio

$\frac{t_1}{t_2}=\frac{\frac{RC}{2}\ln 2}{2RC\ln 2}$

Canceling common terms:

$\frac{t_1}{t_2}=\frac{1}{4}$

5. Evaluate options

• A: $1$ ❌
• B: $\frac{1}{2}$ ❌
• C: $\frac{1}{4}$ ✅
• D: $2$ ❌

Thus, the correct option is C.