The time in seconds required to produce a potential difference of 20V across a capacitor of 1000`mu` F when it is charged at the steady rate of `200 mu C`/s is

Options

A. 50
B. 100
C. 150
D. 200

Answer

To solve the problem, we need to find the time required to produce a potential difference of 20V across a capacitor of 1000 µF when charged at a steady rate of 200 µC/s.

### Step-by-Step Solution:

1. **Convert the given values to standard units:**
- The capacitance \( C \) is given as \( 1000 \, \mu F \).
\[ C = 1000 \, \mu F = 1000 \times 10^{-6} \, F = 10^{-3} \, F \]
- The potential difference \( V \) is given as \( 20 \, V \).
- The rate of charge \( \frac{dq}{dt} \) is given as \( 200 \, \mu C/s \).
\[ \frac{dq}{dt} = 200 \, \mu C/s = 200 \times 10^{-6} \, C/s = 2 \times 10^{-4} \, C/s \]

2. **Calculate the total charge \( Q \) required to produce the potential difference:**
- Using the formula \( Q = C \times V \):
\[ Q = C \times V = (10^{-3} \, F) \times (20 \, V) = 20 \times 10^{-3} \, C = 2 \times 10^{-2} \, C \]

3. **Calculate the time \( t \) required to charge the capacitor:**
- The time can be calculated using the formula:
\[ t = \frac{Q}{\frac{dq}{dt}} \]
- Substituting the values:
\[ t = \frac{2 \times 10^{-2} \, C}{2 \times 10^{-4} \, C/s} = \frac{2}{2} \times 10^{2} \, s = 10^{2} \, s = 100 \, s \]

4. **Conclusion:**
- The time required to produce a potential difference of 20V across the capacitor is \( 100 \, seconds \).

### Final Answer:
The time required is **100 seconds**.