A uniformly wound solenoidal coil of self-inductance 1.8 × 10 ^ - 4...

$\dfrac{1}{Lp} = \dfrac{1}{L} + \dfrac{1}{L} = \dfrac{2}{L} ⇒ {L}_{P} = \dfrac{L}{2}$

Where $L$ is inductance of each part,

$= \dfrac{\text{1.8} \times {10}^{- 4}}{2} = \text{0.9} \times {10}^{- 4}H$

$∴ {L}_{P} = \dfrac{L}{2} = \dfrac{\text{0.9} \times {10}^{- 4}}{2} = \text{0.45} \times {10}^{- 4}H$

Resistance of each part, $r = \dfrac{6}{2} = 3Ω$

Now, $\dfrac{1}{rP} = \dfrac{1}{3} + \dfrac{1}{3} = \dfrac{2}{3}$

Time constant of circuit,

$= \dfrac{LP}{rP} = \dfrac{\text{0.45} \times {10}^{- 4}}{1.5} = 3 \times {10}^{- 5}s$