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,
(iv) Energy is being delivered by the source.

Options

A. `16(1-e^(-1//3))mu J//s`
B. `16 mu J//s`
C. `(16)/(3) e^(-1//3) mu J//s`
D. `(16)/(3) (1-e^(-1//3)) mu J//s`

Answer

To solve the problem of finding the rate at which energy is being delivered by the source in the given circuit, we will follow these steps:

### Step 1: Identify the Components and Parameters
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 \( E = 4 \, V \)

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

### Step 3: Determine the Current at \( t = 1 \, \text{s} \)
The current \( I(t) \) in the circuit at time \( t \) is given by:
\[ I(t) = I_0 e^{-t/\tau} \]
where \( I_0 = \frac{E}{R} \).

First, calculate \( I_0 \):
\[ I_0 = \frac{4}{3 \times 10^6} = \frac{4}{3 \times 10^6} \, A \]

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

### Step 4: Calculate the Rate of Energy Delivery
The power \( P \) delivered by the source is given by:
\[ P = E \cdot I(t) \]
Substituting the values:
\[ P = 4 \cdot \left( \frac{4}{3 \times 10^6} e^{-1/3} \right) \]
\[ P = \frac{16}{3 \times 10^6} e^{-1/3} \, W \]

### Step 5: Numerical Calculation
To find the numerical value, we can use \( e^{-1/3} \approx 0.7165 \):
\[ P \approx \frac{16}{3 \times 10^6} \cdot 0.7165 \approx \frac{11.464}{3 \times 10^6} \approx 3.821 \times 10^{-6} \, W \]

### Final Answer
The rate at which energy is being delivered by the source at \( t = 1 \, \text{s} \) is approximately:
\[ P \approx 3.82 \, \mu W \]