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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.

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.

Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
7 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.

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