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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:6

1 Answers

Navjyot Kalra
askIITians Faculty 654 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 O

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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