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A uniform wound solenoidal coil of self-inductance $1.8 \times 10^{- 4} H$ 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

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$3 \times 10^{- 5} s$

$0.75 \times 10^{- 5} s$

$6 \times 10^{- 5} s$

$1.5 \times 10^{- 5} s$

answer is A.

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Detailed Solution

$\frac{1}{L_{p}} = \frac{1}{L} + \frac{1}{L} = \frac{2}{L} \Rightarrow L_{p} = \frac{L}{2}$

Where $L$ is inductance of each part $= \frac{1.8 \times 10^{- 4}}{2} = 0.9 \times 10^{- 4} H$

$∴ L_{p} = \frac{L}{2} = \frac{0.9 \times 10^{- 4}}{2} = 0.45 \times 10^{- 4} H$

Resistance of each part, $r = \frac{6}{2} = 3 Ω N o w, \frac{1}{r p} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

Time constant of circuit, $= \frac{L_{p}}{r p} = \frac{0.45 \times 10^{- 4}}{1.5} = 3 \times 10^{- 5} s$

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