A capacitor of capacitance `200 muF` is connected to a battery through a resistance of `5 kOmega` at time t = 0. Charge stored in the capacitor in first second in larger than the charge stored in the next

Options

A. 1s
B. 10s
C. 100 s
D. 1,000 s

Explanation

To solve the problem, we need to analyze the charging behavior of a capacitor connected to a resistor and a battery. The key points to consider are the capacitance, resistance, and the time constant of the circuit.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Capacitance (C) = 200 µF = 200 × 10^(-6) F
- Resistance (R) = 5 kΩ = 5000 Ω

2. **Calculate the Time Constant (τ)**:
The time constant τ for an RC circuit is given by the formula:
\[ τ = R \times C \]
Substituting the values:
\[ τ = 5000 \, \Omega \times 200 \times 10^{-6} \, F = 1 \, s \]

3. **Understand the Charging Equation**:
The charge (Q) on a capacitor at time t during charging is given by:
\[ Q(t) = C \cdot V \cdot (1 - e^{-t/τ}) \]
where V is the voltage of the battery and e is the base of the natural logarithm.

4. **Calculate Charge at t = 1 second**:
At t = 1 s, the charge stored in the capacitor is:
\[ Q(1) = C \cdot V \cdot (1 - e^{-1/1}) = C \cdot V \cdot (1 - e^{-1}) \]

5. **Calculate Charge at t = 2 seconds**:
At t = 2 s, the charge stored is:
\[ Q(2) = C \cdot V \cdot (1 - e^{-2/1}) = C \cdot V \cdot (1 - e^{-2}) \]

6. **Compare Charges**:
We need to compare Q(1) and Q(2):
- Since \( e^{-1} \) is approximately 0.3679 and \( e^{-2} \) is approximately 0.1353, we have:
\[ Q(1) = C \cdot V \cdot (1 - 0.3679) = C \cdot V \cdot 0.6321 \]
\[ Q(2) = C \cdot V \cdot (1 - 0.1353) = C \cdot V \cdot 0.8647 \]
- Clearly, \( Q(1) < Q(2) \) since 0.6321 < 0.8647.

7. **Conclusion**:
The charge stored in the capacitor in the first second is less than the charge stored in the next second. Therefore, the statement in the question is incorrect.