A `4 muF` capacitor, a resistance of `2.5 M Omega` is in series with `12V` battery. Find the time after which the potential difference across the capacitor is 3 times the potential difference across the resistor. [ Given In (2) = 0.693]

Options

A. 13.86 s
B. 6.93 s
C. 7 s
D. 14 s

Answer

To solve the problem, we need to find the time \( t \) after which the potential difference across the capacitor \( V_C \) is three times the potential difference across the resistor \( V_R \) in a circuit consisting of a capacitor, a resistor, and a battery.

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance, \( C = 4 \, \mu F = 4 \times 10^{-6} \, F \)
- Resistance, \( R = 2.5 \, M\Omega = 2.5 \times 10^{6} \, \Omega \)
- Voltage of the battery, \( E = 12 \, V \)

2. **Understand the relationships:**
- The potential difference across the capacitor is given by:
\[ V_C = \frac{Q}{C} \]
- The potential difference across the resistor is given by:
\[ V_R = I \cdot R \]
- The current \( I \) in a charging capacitor circuit is given by:
\[ I = \frac{E - V_C}{R} \]

3. **Set up the equation for the relationship between \( V_C \) and \( V_R \):**
- We need to find the time \( t \) when:
\[ V_C = 3 V_R \]
- Substituting the expressions for \( V_C \) and \( V_R \):
\[ \frac{Q}{C} = 3 \left( \frac{E - V_C}{R} \right) R \]
- Rearranging gives:
\[ Q = 3(E - V_C)C \]

4. **Use the charging equation for the capacitor:**
- The charge \( Q \) on the capacitor at time \( t \) is:
\[ Q = C \cdot E \left(1 - e^{-\frac{t}{\tau}}\right) \]
- Where \( \tau = R \cdot C \) is the time constant.

5. **Substituting \( Q \) into the equation:**
- Substitute \( Q \) into the equation:
\[ C \cdot E \left(1 - e^{-\frac{t}{\tau}}\right) = 3(E - \frac{Q}{C})C \]
- This simplifies to:
\[ 1 - e^{-\frac{t}{\tau}} = 3 \left(1 - e^{-\frac{t}{\tau}}\right) \]

6. **Rearranging the equation:**
- Rearranging gives:
\[ 4 e^{-\frac{t}{\tau}} = 1 \]
- Taking the natural logarithm of both sides:
\[ -\frac{t}{\tau} = \ln(0.25) = -2 \ln(2) \]
- Therefore:
\[ t = 2 \tau \ln(2) \]

7. **Calculate \( \tau \):**
- Calculate \( \tau \):
\[ \tau = R \cdot C = (2.5 \times 10^6) \cdot (4 \times 10^{-6}) = 10 \]

8. **Final calculation of time \( t \):**
- Substitute \( \tau \) into the equation for \( t \):
\[ t = 2 \cdot 10 \cdot 0.693 = 13.86 \, seconds \]

### Final Answer:
The time after which the potential difference across the capacitor is three times the potential difference across the resistor is approximately **13.86 seconds**.