A 3 mega ohm resistor and an uncharged `1 mu F` capacitor are connected in a single loop circuit with a constant source of 4 volt. At one second after the connection is made what are the rates at which,
(ii) Energy is being stored in the capacitor.

Options

A. `(16)/(3) (1-e^(-1//3))e^(-1//3) mu J //s`
B. `(16)/(3) (1-e^(-2//3)) mu J//s`
C. `(16)/(3) e^(-2//3) mu J//s`
D. None of these

Answer

To solve the problem step by step, we need to find the rate at which energy is being stored in the capacitor at one second after the connection is made.

### Step 1: Understand the circuit components
We have:
- A resistor \( R = 3 \, \text{M}\Omega = 3 \times 10^6 \, \Omega \)
- A capacitor \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)
- A voltage source \( V = 4 \, V \)

### Step 2: Calculate the time constant \( \tau \)
The time constant \( \tau \) for an RC circuit is given by:
\[ \tau = R \cdot C \]
Substituting the values:
\[ \tau = (3 \times 10^6 \, \Omega) \cdot (1 \times 10^{-6} \, F) = 3 \, \text{s} \]

### Step 3: Determine the charge on the capacitor at time \( t = 1 \, s \)
The charge \( Q(t) \) on the capacitor at any time \( t \) is given by:
\[ Q(t) = Q_{\text{max}} \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( Q_{\text{max}} = C \cdot V \).

Calculating \( Q_{\text{max}} \):
\[ Q_{\text{max}} = (1 \times 10^{-6} \, F) \cdot (4 \, V) = 4 \times 10^{-6} \, C \]

Now substituting \( t = 1 \, s \) and \( \tau = 3 \, s \):
\[ Q(1) = 4 \times 10^{-6} \left(1 - e^{-\frac{1}{3}}\right) \]

### Step 4: Calculate \( \frac{dQ}{dt} \)
The rate of change of charge with respect to time is given by:
\[ \frac{dQ}{dt} = \frac{Q_{\text{max}}}{\tau} e^{-\frac{t}{\tau}} \]
Substituting \( Q_{\text{max}} \) and \( \tau \):
\[ \frac{dQ}{dt} = \frac{4 \times 10^{-6}}{3} e^{-\frac{1}{3}} \]

### Step 5: Calculate the rate of energy storage in the capacitor
The energy \( U \) stored in the capacitor is given by:
\[ U = \frac{1}{2} \frac{Q^2}{C} \]
The rate of energy storage is:
\[ \frac{dU}{dt} = \frac{Q}{C} \frac{dQ}{dt} \]

Substituting \( Q(1) \) and \( \frac{dQ}{dt} \):
\[ \frac{dU}{dt} = \frac{Q(1)}{C} \cdot \frac{dQ}{dt} \]
Now substituting the values we calculated:
\[ \frac{dU}{dt} = \frac{4 \times 10^{-6} \left(1 - e^{-\frac{1}{3}}\right)}{1 \times 10^{-6}} \cdot \frac{4 \times 10^{-6}}{3} e^{-\frac{1}{3}} \]

### Step 6: Simplify the expression
\[ \frac{dU}{dt} = 4 \left(1 - e^{-\frac{1}{3}}\right) \cdot \frac{4}{3} e^{-\frac{1}{3}} = \frac{16}{3} \left(1 - e^{-\frac{1}{3}}\right) e^{-\frac{1}{3}} \, \text{microjoules/second} \]

### Final Answer
The rate at which energy is being stored in the capacitor at \( t = 1 \, s \) is:
\[ \frac{dU}{dt} = \frac{16}{3} \left(1 - e^{-\frac{1}{3}}\right) e^{-\frac{1}{3}} \, \mu J/s \]