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  • A uniformly wound solenoid coil of self inductance  henry and resistance 6

A uniformly wound solenoid coil of self inductance 1.8 \times 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-voll battery of negligible resistance. The time constant for the current in the circuit and the steady state current through the battery is

Option: 1

\mathrm{0.3 \times 10^{-4} \mathrm{~S}, 8 \mathrm{~A}}


Option: 2

1.3 \times 10^{-4} \mathrm{~S}, 8 \mathrm{~A}


Option: 3

0.3 \times 10^{-4} \mathrm{~S}, 5 \mathrm{~A}


Option: 4

none of these


Answers (1)

best_answer

The coil is broken into two identical coils.

\mathrm{\begin{aligned} & L e q=\frac{L / 2 \times L / 2}{L / 2+L / 2}=\frac{L}{4}=0.45 \times 10^{-4} H \\ & \text { Req }=\frac{R / 2 \times R / 2}{R / 2+R / 2}=\frac{R}{4}=1.5 \Omega \end{aligned}}

\mathrm{\text { Time constant }=\frac{\text { Leq. }}{\text { Req. }}=\frac{0.45 \times 10^{-4}}{1.5}=0.3 \times 10^{-4} \mathrm{~s}}

\mathrm{\text { Steady current } \mathrm{I}=\frac{E}{R}=\frac{12}{1.5}=8 \mathrm{~A} \text {. }}

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Ritika Jonwal

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