A charged capacitor of unknown capacitance is connected in series with `100kOmega` resistance and an ideal ammeter. The initial current in the circuit is found to be 0.2mA and the current after 7 sec is found to be 0.1mA. The potential energy that was stored in the capacitor before it was connected in the circuit is ______mJ. [Take `log_(2)2=0.7`]

Explanation

To solve the problem step-by-step, we will follow the outlined process:

### Step 1: Understand the given information
We have:
- Initial current \( I_0 = 0.2 \, \text{mA} = 0.2 \times 10^{-3} \, \text{A} \)
- Current after 7 seconds \( I = 0.1 \, \text{mA} = 0.1 \times 10^{-3} \, \text{A} \)
- Resistance \( R = 100 \, \text{k}\Omega = 100 \times 10^{3} \, \Omega \)

### Step 2: Use Ohm's Law to find the initial voltage across the capacitor
Using Ohm's Law, the initial current can be expressed as:
\[ I_0 = \frac{V}{R} \]
Rearranging gives:
\[ V = I_0 \times R \]
Substituting the values:
\[ V = 0.2 \times 10^{-3} \, \text{A} \times 100 \times 10^{3} \, \Omega = 20 \, \text{V} \]

### Step 3: Determine the time constant and half-life
The current in an RC circuit decays exponentially. The time constant \( \tau \) is given by:
\[ \tau = R \times C \]
The current after time \( t \) is given by:
\[ I(t) = I_0 e^{-t/\tau} \]
Given that the current reduces to half after a certain time, we can set up the equation:
\[ 0.1 \, \text{mA} = 0.2 \, \text{mA} \cdot e^{-7/\tau} \]
Dividing both sides by \( 0.2 \, \text{mA} \):
\[ 0.5 = e^{-7/\tau} \]
Taking the natural logarithm of both sides:
\[ \ln(0.5) = -\frac{7}{\tau} \]
Since \( \ln(0.5) = -\ln(2) \), we have:
\[ -\ln(2) = -\frac{7}{\tau} \implies \tau = \frac{7}{\ln(2)} \]

### Step 4: Calculate the capacitance
Using the relationship \( \tau = R \times C \):
\[ C = \frac{\tau}{R} = \frac{7}{\ln(2) \times 100 \times 10^{3}} \]
Using \( \ln(2) \approx 0.7 \):
\[ C = \frac{7}{0.7 \times 100 \times 10^{3}} = \frac{7}{70 \times 10^{3}} = 100 \, \mu\text{F} \]

### Step 5: Calculate the potential energy stored in the capacitor
The potential energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
Substituting the values:
\[ U = \frac{1}{2} \times 100 \times 10^{-6} \, \text{F} \times (20 \, \text{V})^2 \]
Calculating:
\[ U = \frac{1}{2} \times 100 \times 10^{-6} \times 400 = 20 \times 10^{-3} \, \text{J} = 20 \, \text{mJ} \]

### Final Answer
The potential energy that was stored in the capacitor before it was connected in the circuit is **20 mJ**.