A 8 `mu` F capacitor is charged by a 400 V supply through 0.1 M `Omega` resistance. The time taken by the capacitor to develop a potential difference of 300V is : (Given `log_(10)4 = 0.602`)

Options

A. 2.2 sec
B. 1.1 sec
C. 0.55 sec
D. 0.48 sec

Answer

To solve the problem, we need to determine the time taken by an 8 µF capacitor to develop a potential difference of 300 V when charged by a 400 V supply through a 0.1 MΩ resistor. We can use the formula for the charging of a capacitor in an RC circuit.

### Step-by-step Solution:

1. **Identify the given values:**
- Capacitance, \( C = 8 \, \mu F = 8 \times 10^{-6} \, F \)
- Supply voltage, \( V = 400 \, V \)
- Resistance, \( R = 0.1 \, M\Omega = 0.1 \times 10^{6} \, \Omega = 100,000 \, \Omega \)
- Desired voltage across the capacitor, \( V_c = 300 \, V \)

2. **Use the formula for the voltage across a charging capacitor:**
The voltage across a charging capacitor at time \( t \) is given by:
\[ V_c = V(1 - e^{-\frac{t}{RC}}) \]
Where:
- \( V \) is the supply voltage,
- \( R \) is the resistance,
- \( C \) is the capacitance,
- \( e \) is the base of the natural logarithm.

3. **Rearranging the formula to solve for time \( t \):**
We can rearrange the formula to find \( t \):
\[ 1 - \frac{V_c}{V} = e^{-\frac{t}{RC}} \]
Taking the natural logarithm on both sides:
\[ \ln\left(1 - \frac{V_c}{V}\right) = -\frac{t}{RC} \]
Therefore:
\[ t = -RC \cdot \ln\left(1 - \frac{V_c}{V}\right) \]

4. **Calculate \( 1 - \frac{V_c}{V} \):**
\[ 1 - \frac{300}{400} = 1 - 0.75 = 0.25 \]

5. **Substituting values into the equation:**
Now we can substitute \( R \), \( C \), and \( \ln(0.25) \) into the equation. First, we need to find \( \ln(0.25) \):
\[ \ln(0.25) = \ln\left(\frac{1}{4}\right) = -\ln(4) \]
Given \( \log_{10}(4) = 0.602 \), we can convert this to natural logarithm:
\[ \ln(4) = 0.602 \times 2.303 \approx 1.386 \]
Therefore:
\[ \ln(0.25) = -1.386 \]

6. **Calculate \( t \):**
Now substituting the values:
\[ t = -RC \cdot (-1.386) = RC \cdot 1.386 \]
\[ t = (100,000 \, \Omega)(8 \times 10^{-6} \, F) \cdot 1.386 \]
\[ t = 0.8 \cdot 1.386 \approx 1.109 \, s \]

7. **Final Result:**
Therefore, the time taken by the capacitor to develop a potential difference of 300 V is approximately:
\[ t \approx 1.1 \, s \]