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. 10 J/s
B. 20 J/s
C. 40 J/s
D. 5 J/s

Answer

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

### Step 1: Understand the Circuit
We have a capacitor of capacitance \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \) connected to a battery of emf \( V = 20 \, V \) through a resistor \( R = 10 \, \Omega \). The capacitor starts uncharged.

### Step 2: Determine the Maximum Charge on the Capacitor
The maximum charge \( Q_{\text{max}} \) that the capacitor can hold when fully charged is given by:
\[ Q_{\text{max}} = C \cdot V \]
Substituting the values:
\[ Q_{\text{max}} = (100 \times 10^{-6}) \cdot 20 = 2 \times 10^{-3} \, C = 2 \, mC \]

### Step 3: Energy Stored in the Capacitor
The energy \( U \) stored in a capacitor at any charge \( Q \) is given by:
\[ U = \frac{Q^2}{2C} \]

### Step 4: Charge as a Function of Time
The charge \( Q(t) \) on the capacitor as a function of time \( t \) is given by:
\[ Q(t) = Q_{\text{max}} \left(1 - e^{-t/(RC)}\right) \]
Substituting for \( Q_{\text{max}} \):
\[ Q(t) = 2 \times 10^{-3} \left(1 - e^{-t/(RC)}\right) \]

### Step 5: Differentiate Energy with Respect to Time
To find the rate of change of energy with respect to time, we first substitute \( Q(t) \) into the energy equation:
\[ U(t) = \frac{(Q(t))^2}{2C} = \frac{(2 \times 10^{-3} (1 - e^{-t/(RC)}))^2}{2C} \]
Now, we differentiate \( U(t) \) with respect to \( t \):
\[ \frac{dU}{dt} = \frac{d}{dt} \left( \frac{(2 \times 10^{-3})^2 (1 - e^{-t/(RC)})^2}{2C} \right) \]

### Step 6: Apply the Chain Rule
Using the chain rule and product rule, we can find \( \frac{dU}{dt} \):
\[ \frac{dU}{dt} = \frac{(2 \times 10^{-3})^2}{2C} \cdot 2(1 - e^{-t/(RC)}) \cdot \frac{d}{dt}(1 - e^{-t/(RC)}) \]
The derivative \( \frac{d}{dt}(1 - e^{-t/(RC)}) = \frac{1}{RC} e^{-t/(RC)} \).

### Step 7: Set the Maximum Rate of Energy Storage
To find the maximum rate, we set the derivative equal to zero. This occurs when \( 1 - e^{-t/(RC)} = \frac{1}{2} \), or \( e^{-t/(RC)} = \frac{1}{2} \).

### Step 8: Substitute Back to Find Maximum Rate
The maximum rate of energy storage can be expressed as:
\[ \frac{dU}{dt} = \frac{V^2}{4R} \]
Substituting the values:
\[ \frac{dU}{dt} = \frac{20^2}{4 \times 10} = \frac{400}{40} = 10 \, \text{W} \]

### Final Answer
Thus, the maximum rate at which energy is stored in the capacitor is:
\[ \boxed{10 \, \text{J/s}} \]