The magnetic field inside a 2mH inductor becomes 0-8 of its maximum value in 20- s when the inductor is joined to battery- Find resistance of the circuit-

The correct option is D 160 Ω i∝ B . ∴ current also becomes 0.8 of its maximum value t = τlog_e i_0i_0 i or 20 × 10^ 6 = 2 × ...

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Byju's Answer

Standard XII

Physics

Alternating Current Applied to an Inductor

The magnetic ...

Question

The magnetic field inside a $2 m H$ inductor becomes $0.8$ of its maximum value in $20 μ s$ when the inductor is joined to battery. Find resistance of the circuit.

160 Ω

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80 Ω

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320 Ω

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240 Ω

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none of these

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Solution

The correct option is D $160 Ω$
$i \propto B$ . $∴$ current also becomes $0.8$ of its maximum value

$t = τ {log}_{e} \frac{i_{0}}{i_{0} - i}$

or $20 \times 10^{- 6} = \frac{2 \times 10^{- 3}}{R} {log}_{e} 5$

Resistance of circuit $R = 100 \times 2.303 (0.6990)$
$= 160 Ω$

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The magnetic field inside a 2mH inductor becomes 0.8 of its maximum value in 20μs when the inductor is joined to battery. Find resistance of the circuit.

The correct option is D 160Ωi∝B . ∴ current also becomes 0.8 of its maximum valuet=τlogei0i0−ior 20×10−6=2×10−3Rloge5Resistance of circuit R=100×2.303(0.6990)=160Ω

The magnetic field inside a $2 m H$ inductor becomes $0.8$ of its maximum value in $20 μ s$ when the inductor is joined to battery. Find resistance of the circuit.

The correct option is D $160 Ω$
$i \propto B$ . $∴$ current also becomes $0.8$ of its maximum value

$t = τ {log}_{e} \frac{i_{0}}{i_{0} - i}$

or $20 \times 10^{- 6} = \frac{2 \times 10^{- 3}}{R} {log}_{e} 5$

Resistance of circuit $R = 100 \times 2.303 (0.6990)$
$= 160 Ω$