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PhysicsEMI/ACSelf and mutual inductorMedium2 minPYQ_2020

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A uniformly 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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Answer:

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, $r = \frac{6}{2} = 3 Ω$

Now, $\frac{1}{r_{P}} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

Time constant of circuit,

$= \frac{L_{P}}{r_{P}} = \frac{0.45 \times 10^{- 4}}{1.5} = 3 \times 10^{- 5} s$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Self and mutual inductor

⏱ 2mℹ️ Source: PYQ_2020

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