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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics Class 12th Electromagnetic Induction

A $10 Ω$ 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $μ s .$ The average e.m.f. induced in the coil is__________V.

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Raj Pandey | Contributor-Level 9

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Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i₀ $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

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Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i₀ $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

= $= - L i_{0} \frac{- 1}{2} e^{- t / τ}$

$e = \frac{L i_{0}}{τ} e^{- t / τ}$

$e_{a v g} = \frac{\int_{0}^{200 μ s} e d t}{\int_{0}^{200 μ s} d t} = \frac{\int_{0}^{200 μ s} \frac{L i_{0}}{τ} e^{- t / τ} d t}{\int_{0}^{200 μ s} d t}$

$= \frac{L i_{0}}{200 \times 1 0^{- 6} s e c} = \frac{20 \times 1 0^{- 3} \times 2}{200 \times 1 0^{- 6}} = 400$

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A $10 Ω$ 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $μ s .$ The average e.m.f. induced in the coil is__________V.

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A 1 0 Ω 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 μ s . The average e.m.f. induced in the coil is__________V.

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A $10 Ω$ 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $μ s .$ The average e.m.f. induced in the coil is__________V.