A uniformly wound solenoidal coil of self-inductance 1.8 × 10 - 4 H and resistance 6 Ω is broken up into two identical coils. These identical coils are then connected in parallel across a 12 V battery of negligible resistance. The time constant of the circuit is

Step-by-step solution

1 L p = 1 L + 1 L = 2 L         ⇒ L P = L 2   Where L is inductance of each part, =   1.8   × 10 - 4 2 = 0.9 × 10 - 4   H ∴   L P = L 2 =   0.9 × 10 - 4 2 = 0.45 × 10 - 4   H Resistance of each part, r = 6 2 = 3   Ω Now, 1 r P = 1 3 + 1 3 = 2 3 Time constant of circuit, = L P r P = 0.45 × 10 - 4 1.5 = 3 × 10 - 5   s