A `16 muF` capacitor , initially charged to 5 V, is started charging at t=0 by a source at the rate of `40 tmuCs^(-1)`. How long will it take to raise its potential to 10 V?

Options

A. 1 s
B. 2 s
C. 3 s
D. none of these

Explanation

To solve the problem, we need to determine how long it will take for a `16 µF` capacitor, initially charged to `5 V`, to reach `10 V` when charged at a rate of `40 µC/s`.

### Step-by-step Solution:

1. **Calculate Initial Charge (Qi)**:
The initial charge on the capacitor can be calculated using the formula:
\[ Q_i = C \times V_i \]
Where:
- \( C = 16 \, \mu F = 16 \times 10^{-6} \, F \)
- \( V_i = 5 \, V \)

Substituting the values:
\[ Q_i = 16 \times 10^{-6} \, F \times 5 \, V = 80 \, \mu C \]

2. **Calculate Final Charge (Qf)**:
The final charge when the capacitor reaches `10 V` is:
\[ Q_f = C \times V_f \]
Where:
- \( V_f = 10 \, V \)

Substituting the values:
\[ Q_f = 16 \times 10^{-6} \, F \times 10 \, V = 160 \, \mu C \]

3. **Determine the Change in Charge (ΔQ)**:
The change in charge required to go from \( Q_i \) to \( Q_f \) is:
\[ \Delta Q = Q_f - Q_i = 160 \, \mu C - 80 \, \mu C = 80 \, \mu C \]

4. **Rate of Charge Flow**:
The rate at which the capacitor is charged is given as:
\[ \frac{dQ}{dt} = 40 \, \mu C/s \]

5. **Calculate Time (t)**:
To find the time required to charge the capacitor from \( Q_i \) to \( Q_f \), we can use the relationship:
\[ \Delta Q = \frac{dQ}{dt} \times t \]
Rearranging for \( t \):
\[ t = \frac{\Delta Q}{\frac{dQ}{dt}} = \frac{80 \, \mu C}{40 \, \mu C/s} = 2 \, s \]

### Final Answer:
The time required to raise the potential of the capacitor from `5 V` to `10 V` is **2 seconds**.

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