SATHEE: Electromagnetic Induction and Alternating Current 4 Question 15

Electromagnetic Induction and Alternating Current 4 Question 15

####15. 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 $15 V$ battery of negligible resistance. The time constant for the current in the circuit is …… s and the steady state current through the battery is … A. (1989, 2M)

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Answer:

Correct Answer: 15. $3 \times 10^{- 5}, 10$

Solution:

Inductance of the circuit $L = \frac{0.9 \times 10^{- 4}}{2} = 0.45 \times 10^{- 4} H$ (in parallel)

Resistance of the circuit $R = 3 / 2 = 1.5 Ω$ (in parallel)

$∴ τ_{L} ($ time constant $) = \frac{L}{R} = 3.0 \times 10^{- 5} s$

Steady state current in the circuit through the battery

$i_{0} = \frac{V}{R} = \frac{15}{1.5} = 10 A$

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Electromagnetic Induction and Alternating Current 4 Question 15

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