An uncharged capacitor of capacitance `100 mu F` is connected to a battery of emf 20V at t = 0 through a resistance `10 Omega`, then
(i) the maximum rate at which energy is stored in the capacitor is :

Options

A. (4 ln 2) ms
B. (2 ln 2) ms
C. (ln 2) ms
D. (3 ln 2) ms

Answer

To solve the problem of finding the maximum rate at which energy is stored in a capacitor, we can follow these steps:

### Step 1: Understand the energy stored in a capacitor
The energy \( U \) stored in a capacitor is given by the formula:
\[ U = \frac{Q^2}{2C} \]
where \( Q \) is the charge on the capacitor and \( C \) is the capacitance.

### Step 2: Differentiate the energy with respect to time
To find the rate at which energy is being stored, we differentiate \( U \) with respect to time \( t \):
\[ \frac{dU}{dt} = \frac{d}{dt} \left( \frac{Q^2}{2C} \right) = \frac{1}{2C} \cdot 2Q \frac{dQ}{dt} = \frac{Q}{C} \frac{dQ}{dt} \]
Here, \( \frac{dQ}{dt} \) is the current \( I \) flowing into the capacitor.

### Step 3: Substitute the expression for current
In an RC circuit, the charge \( Q \) on the capacitor as a function of time is given by:
\[ Q(t) = C \cdot V \left(1 - e^{-\frac{t}{RC}}\right) \]
where \( V \) is the voltage of the battery, \( R \) is the resistance, and \( C \) is the capacitance. The current \( I \) can be expressed as:
\[ I(t) = \frac{dQ}{dt} = \frac{V}{R} e^{-\frac{t}{RC}} \]

### Step 4: Find the maximum rate of energy storage
The maximum rate of energy storage occurs when \( \frac{dU}{dt} \) is maximized. From the previous step, we can express this as:
\[ \frac{dU}{dt} = \frac{Q}{C} I \]
Substituting \( I \) into this equation gives:
\[ \frac{dU}{dt} = \frac{Q}{C} \cdot \frac{V}{R} e^{-\frac{t}{RC}} \]

### Step 5: Use the maximum charge
The maximum charge \( Q_{\text{max}} \) on the capacitor when fully charged is:
\[ Q_{\text{max}} = C \cdot V = 100 \times 10^{-6} \, \text{F} \times 20 \, \text{V} = 2 \times 10^{-3} \, \text{C} \]

### Step 6: Substitute values to find maximum power
At maximum energy storage, we can substitute \( Q = Q_{\text{max}} \):
\[ \frac{dU}{dt} = \frac{Q_{\text{max}}}{C} \cdot \frac{V}{R} = \frac{2 \times 10^{-3}}{100 \times 10^{-6}} \cdot \frac{20}{10} \]
Calculating this gives:
\[ \frac{dU}{dt} = \frac{2 \times 10^{-3}}{100 \times 10^{-6}} \cdot 2 = 40 \, \text{W} \]

### Final Answer
The maximum rate at which energy is stored in the capacitor is:
\[ \boxed{40 \, \text{W}} \]