LetC 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. `1//2`
C. `1//4`
D. 2

Explanation

To solve the problem, we need to find the ratio \( \frac{t_1}{t_2} \), where \( t_1 \) 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.

### Step 1: Understand the formulas for energy and charge in a capacitor
The energy \( U \) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
where \( C \) is the capacitance and \( V \) is the voltage across the capacitor.

The charge \( Q \) on a capacitor is given by:
\[ Q = C V \]

### Step 2: Determine the time \( t_1 \) for energy to reduce to half
When the capacitor discharges through a resistor, the voltage across the capacitor decreases exponentially:
\[ V(t) = V_0 e^{-\frac{t}{RC}} \]
where \( V_0 \) is the initial voltage, \( R \) is the resistance, and \( C \) is the capacitance.

The energy at time \( t \) is:
\[ U(t) = \frac{1}{2} C [V(t)]^2 = \frac{1}{2} C [V_0 e^{-\frac{t}{RC}}]^2 = \frac{1}{2} C V_0^2 e^{-\frac{2t}{RC}} \]

To find \( t_1 \), we set \( U(t_1) = \frac{1}{2} U_0 \):
\[ \frac{1}{2} C V_0^2 e^{-\frac{2t_1}{RC}} = \frac{1}{2} \left(\frac{1}{2} C V_0^2\right) \]
This simplifies to:
\[ e^{-\frac{2t_1}{RC}} = \frac{1}{2} \]
Taking the natural logarithm of both sides:
\[ -\frac{2t_1}{RC} = \ln\left(\frac{1}{2}\right) \]
Thus,
\[ t_1 = -\frac{RC}{2} \ln\left(\frac{1}{2}\right) = \frac{RC}{2} \ln(2) \]

### Step 3: Determine the time \( t_2 \) for charge to reduce to one-fourth
The charge on the capacitor also decreases exponentially:
\[ Q(t) = Q_0 e^{-\frac{t}{RC}} \]
where \( Q_0 = C V_0 \).

To find \( t_2 \), we set \( Q(t_2) = \frac{1}{4} Q_0 \):
\[ Q_0 e^{-\frac{t_2}{RC}} = \frac{1}{4} Q_0 \]
This simplifies to:
\[ e^{-\frac{t_2}{RC}} = \frac{1}{4} \]
Taking the natural logarithm of both sides:
\[ -\frac{t_2}{RC} = \ln\left(\frac{1}{4}\right) \]
Thus,
\[ t_2 = -RC \ln\left(\frac{1}{4}\right) = RC \ln(4) \]

### Step 4: Find the ratio \( \frac{t_1}{t_2} \)
Now we can find the ratio:
\[ \frac{t_1}{t_2} = \frac{\frac{RC}{2} \ln(2)}{RC \ln(4)} = \frac{1}{2} \cdot \frac{\ln(2)}{\ln(4)} \]
Since \( \ln(4) = \ln(2^2) = 2 \ln(2) \):
\[ \frac{t_1}{t_2} = \frac{1}{2} \cdot \frac{\ln(2)}{2 \ln(2)} = \frac{1}{4} \]

### Final Answer
The ratio \( \frac{t_1}{t_2} \) is:
\[ \frac{t_1}{t_2} = \frac{1}{4} \]