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Grade 11Electromagnetic Induction

A uniformly wound solenoidal coil of self inductance 1.8 x 10-4 henry and resistance 6 ohm is broken up into two identical coils. These identical coils are then connected in parallel across a 15-volt battery of negligible resistance. The time constant for the current in the circuit is …….. Seconds and the steady state current through the battery is ……… amperes.

Profile image of Radhika Batra
12 Years agoGrade 11
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1 Answer

Profile image of Kevin Nash
12 Years ago
The coil is broken into two identical coils.
Leq = L / 2 X L / 2 / L / 2 + L / 2 = L/4 = 0.45 X 10-4 H,
= R / 2 X R / 2 / R / 2 + R / 2 = R / 4 = 1.5 Ω
Time constant = Leq / Req = 0.45 x 10-4 / 1.5 = 0.3 x 10-4 s
Steady current I = E / Req = 15/ 1.5 = 10 A.