A condenser of capacity `500 mu F` is charged at the rate of `400 mu F` per second. The time required to raise its potential by `40 V` is.

Options

A. 50 s
B. 100 s
C. 20 s
D. 10 s

Explanation

To solve the problem, we need to determine the time required to raise the potential of a capacitor (condenser) with a given capacitance when charged at a specific rate.

### Step-by-Step Solution:

1. **Understand the relationship between charge, capacitance, and voltage**:
The relationship is given by the formula:
\[ Q = C \cdot V \]
where \( Q \) is the charge in coulombs, \( C \) is the capacitance in farads, and \( V \) is the voltage in volts.

2. **Identify the given values**:
- Capacitance, \( C = 500 \, \mu F = 500 \times 10^{-6} \, F \)
- Rate of charging, \( \frac{dQ}{dt} = 400 \, \mu C/s = 400 \times 10^{-6} \, C/s \)
- Desired increase in voltage, \( dV = 40 \, V \)

3. **Relate charge change to voltage change**:
From the formula \( Q = C \cdot V \), we can differentiate both sides with respect to time \( t \):
\[ \frac{dQ}{dt} = C \cdot \frac{dV}{dt} \]

4. **Substitute the known values into the equation**:
\[ 400 \times 10^{-6} = 500 \times 10^{-6} \cdot \frac{dV}{dt} \]

5. **Solve for \( \frac{dV}{dt} \)**:
Rearranging the equation gives:
\[ \frac{dV}{dt} = \frac{400 \times 10^{-6}}{500 \times 10^{-6}} = \frac{400}{500} = 0.8 \, V/s \]

6. **Calculate the time required to raise the potential by 40 V**:
Using the relationship \( dV = \frac{dV}{dt} \cdot dt \):
\[ dt = \frac{dV}{\frac{dV}{dt}} = \frac{40 \, V}{0.8 \, V/s} = 50 \, s \]

### Final Answer:
The time required to raise the potential by 40 V is **50 seconds**.

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