Q. A uniformly wound solenoidal coil of self inductance $1.8 \times 10^{- 4}$ H and resistance $6\ \ \Omega$ 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 current in the circuit and the steady state current through battery is
A) $3\ \times 10^{- 5}$ s, 8 A
B) $1.5\ \times 10^{- 5}$ s, 8 A
C) $0.75\ \times 10^{- 4}$ s, 4 A
D) $6\ \times 10^{- 5}$ s, 2 A
Explanation
(a) $\frac{1}{L_{p}} = \frac{1}{L} + \frac{1}{L} = \frac{2}{L}\ \ \ \ or\ \ \ 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
$\therefore\ L_{P} = \frac{L}{2} = \ \frac{0.9 \times 10^{- 4}}{2} = 0.45 \times 10^{- 4}$H
Resistance of each part, $r = \frac{6}{2} = 3\Omega$
Now, $\frac{1}{r_{P}} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$
$\therefore\ r_{P} = 3/2\Omega$
Time constant of circuit $= \frac{L_{P}}{r_{P}} = \ \frac{0.45 \times 10^{- 4}}{3/2} = 8$A
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.