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. `(1)/(2)`
C. `(1)/(4)`
D. `2`

Answer

To solve the problem, we need to determine the ratio \( \frac{t_1}{t_2} \), where \( t_1 \) is the time taken for the energy stored in a 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-by-step Solution:

1. **Initial Energy of the Capacitor**:
The initial energy \( E_0 \) stored in the capacitor is given by:
\[ E_0 = \frac{Q_0^2}{2C} \]
where \( Q_0 \) is the initial charge and \( C \) is the capacitance.

2. **Energy at Time \( t_1 \)**:
At time \( t_1 \), the energy is half of the initial energy:
\[ E(t_1) = \frac{E_0}{2} = \frac{Q^2}{2C} \]
Setting these equal gives:
\[ \frac{Q^2}{2C} = \frac{Q_0^2}{4C} \]
Simplifying this, we find:
\[ Q^2 = \frac{Q_0^2}{2} \implies Q = \frac{Q_0}{\sqrt{2}} \]

3. **Charge Decay Formula**:
The charge \( Q \) at time \( t \) is given by:
\[ Q = Q_0 e^{-\frac{t}{RC}} \]
Substituting \( Q = \frac{Q_0}{\sqrt{2}} \) at \( t = t_1 \):
\[ \frac{Q_0}{\sqrt{2}} = Q_0 e^{-\frac{t_1}{RC}} \]
Dividing both sides by \( Q_0 \):
\[ \frac{1}{\sqrt{2}} = e^{-\frac{t_1}{RC}} \]

4. **Taking Logarithms**:
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{\sqrt{2}}\right) = -\frac{t_1}{RC} \]
This simplifies to:
\[ -\frac{1}{2} \ln(2) = -\frac{t_1}{RC} \implies t_1 = RC \cdot \frac{1}{2} \ln(2) \]

5. **Charge at Time \( t_2 \)**:
Now, we consider the time \( t_2 \) when the charge reduces to one-fourth of its initial value:
\[ Q(t_2) = \frac{Q_0}{4} \]
Using the charge decay formula:
\[ \frac{Q_0}{4} = Q_0 e^{-\frac{t_2}{RC}} \]
Dividing both sides by \( Q_0 \):
\[ \frac{1}{4} = e^{-\frac{t_2}{RC}} \]

6. **Taking Logarithms Again**:
Taking the natural logarithm:
\[ \ln\left(\frac{1}{4}\right) = -\frac{t_2}{RC} \]
This simplifies to:
\[ -2 \ln(2) = -\frac{t_2}{RC} \implies t_2 = RC \cdot 2 \ln(2) \]

7. **Finding the Ratio \( \frac{t_1}{t_2} \)**:
Now we can find the ratio:
\[ \frac{t_1}{t_2} = \frac{RC \cdot \frac{1}{2} \ln(2)}{RC \cdot 2 \ln(2)} = \frac{\frac{1}{2}}{2} = \frac{1}{4} \]

### Conclusion:
Thus, the ratio \( \frac{t_1}{t_2} \) is \( \frac{1}{4} \).