A uniformly wound solenoid coil of self inductance `1.8xx10^(-4)H` and resistance `6Omega` is broken up into two identical coils. These identical coils are then connected in parallel across a `12V` battery of negligible resistance. The time constant and steady state current will be

Options

A. `10musec,6A`
B. `10musec,8A`
C. `30musec,6A`
D. `30musec,8A`

Explanation

To solve the problem, we need to determine the time constant and the steady-state current for the two identical coils connected in parallel across a 12V battery. Let's break down the solution step by step.

### Step 1: Determine the Inductance of Each Coil
The original solenoid has a self-inductance \( L = 1.8 \times 10^{-4} \, \text{H} \). When the solenoid is cut into two identical coils, the inductance of each coil is halved.

\[ L_1' = \frac{L}{2} = \frac{1.8 \times 10^{-4}}{2} = 9.0 \times 10^{-5} \, \text{H} \]

### Step 2: Calculate the Equivalent Inductance of the Parallel Coils
For two inductors in parallel, the equivalent inductance \( L_{\text{eq}} \) is given by:

\[ \frac{1}{L_{\text{eq}}} = \frac{1}{L_1'} + \frac{1}{L_2'} \]

Since both inductors are identical:

\[ \frac{1}{L_{\text{eq}}} = \frac{1}{9.0 \times 10^{-5}} + \frac{1}{9.0 \times 10^{-5}} = \frac{2}{9.0 \times 10^{-5}} \]

Thus,

\[ L_{\text{eq}} = \frac{9.0 \times 10^{-5}}{2} = 4.5 \times 10^{-5} \, \text{H} \]

### Step 3: Determine the Resistance of Each Coil
The original resistance \( R = 6 \, \Omega \). When the solenoid is cut into two identical coils, the resistance of each coil is also halved:

\[ R_1' = \frac{R}{2} = \frac{6}{2} = 3 \, \Omega \]

### Step 4: Calculate the Equivalent Resistance of the Parallel Coils
For two resistors in parallel, the equivalent resistance \( R_{\text{eq}} \) is given by:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1'} + \frac{1}{R_2'} \]

Since both resistors are identical:

\[ \frac{1}{R_{\text{eq}}} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

Thus,

\[ R_{\text{eq}} = \frac{3}{2} = 1.5 \, \Omega \]

### Step 5: Calculate the Time Constant
The time constant \( \tau \) is given by the formula:

\[ \tau = \frac{L_{\text{eq}}}{R_{\text{eq}}} \]

Substituting the values we found:

\[ \tau = \frac{4.5 \times 10^{-5}}{1.5} = 3.0 \times 10^{-5} \, \text{s} = 30 \, \mu s \]

### Step 6: Calculate the Steady-State Current
The steady-state current \( I \) can be calculated using Ohm's Law:

\[ I = \frac{V}{R_{\text{eq}}} \]

Where \( V = 12 \, \text{V} \):

\[ I = \frac{12}{1.5} = 8 \, \text{A} \]

### Final Answers
- Time Constant \( \tau = 30 \, \mu s \)
- Steady-State Current \( I = 8 \, \text{A} \)

---