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,
(iii) Joule heat is appearing in the resistor.

Options

A. `(16)/(3)e^(-1//3) mu J//s`
B. `(1)/(2) e ^(-1//3) mu J//s`
C. `(16)/(3) (e^(-2//3)) mu J//s`
D. `(16)/(3) (1-e^(-1//3))^(2) mu J//s`

Answer

To solve the problem of finding the rate at which Joule heat is appearing in the resistor at one second after the connection is made, we can follow these steps:

### Step 1: Identify the components and their values
We have:
- Resistor (R) = 3 mega ohms = \(3 \times 10^6 \, \Omega\)
- Capacitor (C) = 1 microfarad = \(1 \times 10^{-6} \, F\)
- Voltage (E) = 4 volts

### Step 2: Write the Kirchhoff's loop law equation
According to Kirchhoff's loop law, we can write the equation for the circuit as:
\[ E - IR - \frac{Q}{C} = 0 \]
Where:
- \(I\) is the current,
- \(Q\) is the charge on the capacitor.

### Step 3: Relate current and charge
The current \(I\) can be expressed as:
\[ I = \frac{dQ}{dt} \]
Substituting this into the loop equation gives:
\[ E - R\frac{dQ}{dt} - \frac{Q}{C} = 0 \]

### Step 4: Rearranging the equation
Rearranging the equation, we get:
\[ R\frac{dQ}{dt} = E - \frac{Q}{C} \]

### Step 5: Solve the differential equation
To solve this, we can separate variables:
\[ \frac{dQ}{E - \frac{Q}{C}} = \frac{dt}{R} \]
Integrating both sides will give us the relationship between charge \(Q\) and time \(t\).

### Step 6: Find the maximum current
The maximum current \(I_0\) in the circuit when the capacitor is uncharged is given by:
\[ I_0 = \frac{E}{R} = \frac{4}{3 \times 10^6} \, A \]

### Step 7: Current as a function of time
The current \(I(t)\) at time \(t\) can be expressed as:
\[ I(t) = I_0 e^{-\frac{t}{RC}} \]
Where \(RC\) is the time constant of the circuit:
\[ RC = (3 \times 10^6) \times (1 \times 10^{-6}) = 3 \, seconds \]
Thus,
\[ I(t) = \frac{4}{3 \times 10^6} e^{-\frac{t}{3}} \]

### Step 8: Calculate the current at \(t = 1\) second
Substituting \(t = 1\) second into the equation:
\[ I(1) = \frac{4}{3 \times 10^6} e^{-\frac{1}{3}} \]

### Step 9: Calculate the power (Joule heat) in the resistor
The power \(P\) (rate of Joule heat) in the resistor is given by:
\[ P = I^2 R \]
Substituting \(I(1)\) into this equation:
\[ P = \left(\frac{4}{3 \times 10^6} e^{-\frac{1}{3}}\right)^2 \times (3 \times 10^6) \]

### Step 10: Simplify the expression
Calculating this gives:
\[ P = \frac{16}{(3 \times 10^6)^2} e^{-\frac{2}{3}} \times (3 \times 10^6) \]
\[ P = \frac{16}{3 \times 10^6} e^{-\frac{2}{3}} \]

### Final Result
The rate at which Joule heat is appearing in the resistor at \(t = 1\) second is:
\[ P = \frac{16}{3} e^{-\frac{2}{3}} \, \text{watts} \]