Question

A 20 Henry inductor coil is connected to a 10-ohm resistance in series as shown in the figure. The time at which the rate of dissipation of energy (Joule's heat) across the resistance is equal to the rate at which magnetic energy is stored in the inductor, is ______.

Options

• `2/("In"  2)`

• `1/2` In 2

• 2 In 2

• In 2

Answer 1

A 20 Henry inductor coil is connected to a 10-ohm resistance in series as shown in the figure. The time at which the rate of dissipation of energy (Joule's heat) across the resistance is equal to the rate at which magnetic energy is stored in the inductor, is 2 In 2.

Explanation:

Given: For the LR circuit depicted below, L = 20H, R = 10Ω.

To find: t', the time at which the rate of energy dissipation across the resistance equals the rate of magnetic energy stored in the inductor.

Current flowing through the LR circuit:

`i = E/R(1 - e^{-Rt"/"L})` ...........(i)

The following energy is stored in the inductor:

`U_L = 1/2Li^2`

Rate of energy storage in the inductor:

`(dU_L)/(dt) = 1/2L(2i)(di)/(dt) = Li(di)/(dt)` .......(ii)

The rate at which the resistor dissipates energy:

`(dU_R)/(dt) = i^2R` ..........(iii)

At t = t':

`(dU_L)/(dt) = (dU_R)/(dt)`

`Li(di)/(dt) = i^2R`

`(di)/(dt) = R/Li` .....(iv)

Enter the value of i from the equation (i),

`d/dt[E/R(1 - e^{-Rt"/"L})] = R/L xx E/R(1 - e^{-Rt"/"L})`

Put t = t'.

`R/L e^{-Rt^'"/"L} = R/L(1 - e^{-Rt^'"/"L})`

`e^{-Rt^'"/"L} = (1 - e^{-Rt^'"/"L})`

`e^{-Rt^'"/"L} = 1/2`

`(Rt^')/L` = In 2

t' = `L/R` In 2

t' = `20/10` In 2 = 2 In 2