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

Question

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, 3 × 10 - 5 s, 1.5 × 10 - 5 s, 0.75 × 10 - 5 s, 6 × 10 - 5 s

MARKS App · Accepted Answer

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

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