Question

A coil an inductance of 53 mH and a resistance of .35 ohm If a12 V emf is applied across the coil how much energy is stored in the magnetic field after the current

Answer 1

L = 53mH = 53 × 10^-3 H

resistance , R = 0.35Ω

emf (electromotive force) of battery, V = 12 volts.

current through inductorI = V/R

                                           = (12/0.35) = 34.28 A

now, energy stored in the magnetic field after the current has built up to its equilibrium, E = 1/2 LI²

= 1/2 × 53 × 10^-3 × (34.28)²

≈ 31 J

Answer 2

Answer:

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Solution:-

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Given-

• inductance (L)=53Mh. =53 x 10^-3h
• Resistance =35Ω
• electromotive force (emf)=12V

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To find :-

energy=?

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Formula to be used :-

• I=V/R
• E=1/2LI^2

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first we find current,

So ,

I=V/R

=>12/35

=>0.3428Ampere

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Now ,energy stored is

E=1/2LI^2

=>1/2 x 53 x10^-3 x(0.3428)^2

=>-3114.06J

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hence we get answer ✊