A uniformly wound solenoid coil of slf inductnace 4 mH and resistance `12 Omega` is broken int two parts. These two coil are connected in parallel across is `12V` battery. The steady current through battery is

Options

A. 1A
B. 2A
C. 4A
D. 8A

Explanation

To solve the problem, we will follow these steps:

1. **Understand the Components**: We have a solenoid coil with a self-inductance of 4 mH and a resistance of 12 Ω. When this coil is broken into two parts, each part will have half the inductance and half the resistance.

2. **Determine the New Values**:
- The self-inductance of each part will be \( \frac{4 \text{ mH}}{2} = 2 \text{ mH} \).
- The resistance of each part will be \( \frac{12 \text{ Ω}}{2} = 6 \text{ Ω} \).

3. **Connect in Parallel**: When the two coils are connected in parallel, the total resistance \( R_{total} \) can be calculated using the formula for resistors in parallel:
\[ \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \]
where \( R_1 = 6 \text{ Ω} \) and \( R_2 = 6 \text{ Ω} \).

4. **Calculate the Total Resistance**:
\[ \frac{1}{R_{total}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]
Therefore, \( R_{total} = 3 \text{ Ω} \).

5. **Apply Ohm's Law**: The steady current \( I \) through the battery can be calculated using Ohm's Law:
\[ I = \frac{V}{R_{total}} \]
where \( V = 12 \text{ V} \).

6. **Calculate the Current**:
\[ I = \frac{12 \text{ V}}{3 \text{ Ω}} = 4 \text{ A} \]

Thus, the steady current through the battery is **4 A**.