Question

$\huge\underline{\bf \orange{Question :}}$

An inductance l and a resistance r are connected in series with a battery of emf the maximum rate at which the energy is stored in the magnetic field is​

Answer 1

$\huge\underline{\bf \orange{Solution :}}$

An inductance L and a resistance R are connected in series with a battery of EMF E. The maximum rate at which the energy stored in the magnetic field is P, then what is E=?

When series RL circuit is excited by step voltage, current is i = E/R(1-exp(-tR/L))

di/dt = E/L exp(-tR/L)

Energy J = 1/2 Li^2

dJ/dt = Li di/dt = LE/R(1-exp(-tR/L)) E/L exp(-tR/L)

At t= 0, dJ/dt = 0, at t = infinity dJ/dt = 0

To find maxima, find second derivative. After simplification,

dJ/dt = E^2/R(exp(-tR/L - exp(-2tR/L))……[1]

d2J/dt2 = E^2/R(-R/L exp(-tR/L) + 2R/L(exp(-2tR/L)) = 0

so R/L exp(-tR/L) = 2R/L(exp(-2tR/L)

1/2 = exp(-tR/L)

tR/L = ln 2

t = L/R ln 2

Substitute in [1] to get max value P

P = E^2/R(exp(-L/R ln2 R/L) - exp(-2 L/R ln 2 R/L))

= E^2/R(exp(ln2)^(-1) - exp(ln2)^-2)

= E^2/R(1/2–1/4)

= E^2/(R/4)

so E = 1/2 sqrt(PR)

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Answer 2

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L and a resistance R are connected in series with a battery of emf ε The maximum rate at which the energy is stored in the magnetic field is.