A uniform wound solenoidal coil of self inductance `1.8xx10^(-4)` henry and resistance 6 ohm is broken up into two identical coils. These identical coils are then connected in parallel across a 15-volt battery of negligible resistance. The time constant for the current in the circuit is.......seconds and the steady state current through the battery is ..............amperes.

Explanation

To solve the problem step by step, we will break it down into manageable parts:

### Step 1: Understand the Problem
We have a solenoidal coil with a self-inductance \( L = 1.8 \times 10^{-4} \) H and a resistance \( R = 6 \) ohms. This coil is broken into two identical coils, which are then connected in parallel across a 15-volt battery.

### Step 2: Calculate the Self-Inductance of Each Identical Coil
When the original coil is broken into two identical coils, the self-inductance of each coil becomes:
\[
...