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

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JEE Mains Previous Paper 1 (Held On: 08 Apr 2019 Shift 1)

Option 3 : 2 ln 2

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

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**Concept: **

The power is given by the formula:

P = i^{2} R

The relationship between the power and the inductance is:

\({\rm{P}} = \left[ {{\rm{L}}\frac{{{\rm{di}}}}{{{\rm{dt}}}}} \right]{\rm{I}}\)

On equating both the power,

\(\Rightarrow \left[ {{\rm{L}}\frac{{{\rm{di}}}}{{{\rm{dt}}}}} \right]{\rm{i}} = {{\rm{i}}^2}{\rm{R}}\)

\(\Rightarrow \left[ {{\rm{L}}\frac{{{\rm{di}}}}{{{\rm{dt}}}}} \right] = {\rm{iR}}\)

\(\Rightarrow \frac{{{\rm{di}}}}{{{\rm{dt}}}} = \frac{{{\rm{iR}}}}{{\rm{L}}}\)

We know that, the time constant in LR circuit is given by the formula:

\({\rm{\tau }} = \frac{{\rm{L}}}{{\rm{R}}}\)

Now, the equation becomes,

\(\Rightarrow \frac{{{\rm{di}}}}{{{\rm{dt}}}} = \frac{{\rm{i}}}{{\rm{\tau }}}\)

Now, the change in current and its rate is given as:

I = I_{f}(1-e^{(-t/τ)}) ----(1)

On differentiating,

\(\frac{{{\rm{dI}}}}{{{\rm{dt}}}} = {{\rm{I}}_{\rm{f}}}\frac{{\rm{d}}}{{{\rm{dt}}}}\left( {1 - {{\rm{e}}^{ - {\rm{t}}/{\rm{\tau }}}}} \right) = - {{\rm{I}}_{\rm{f}}}{{\rm{e}}^{ - {\rm{t}}/{\rm{\tau }}}}\left( { - \frac{1}{{\rm{\tau }}}} \right)\)

\(\therefore \frac{{{\rm{dI}}}}{{{\rm{dt}}}} = \frac{{{{\rm{I}}_{\rm{f}}}}}{{\rm{\tau }}}{{\rm{e}}^{ - {\rm{t}}/{\rm{\tau }}}}\)

The relationship between I and time constant is given as:

\({\rm{I}} = {\rm{\tau }}\frac{{{\rm{dI}}}}{{{\rm{dt}}}}\) ----(2)

On equating the equations (1) and (2),

\(\Rightarrow {{\rm{I}}_{\rm{f}}}\left( {1 - {{\rm{e}}^{ - {\rm{t}}/{\rm{\tau }}}}} \right) = {\rm{\tau }}\frac{{{\rm{dI}}}}{{{\rm{dt}}}}\)

Now substituting the rate of change of current,

\(\Rightarrow {{\rm{I}}_{\rm{f}}}\left( {1 - {{\rm{e}}^{ - {\rm{t}}/{\rm{\tau }}}}} \right) = {\rm{\tau }}\left( {\frac{{{{\rm{I}}_{\rm{f}}}}}{{\rm{\tau }}}{{\rm{e}}^{ - {\rm{t}}/{\rm{\tau }}}}} \right)\)

⇒ 1 - e^{(-t/τ) }= e^{(-t/τ)}

⇒ 1 = 2e^{(-t/τ)}

⇒ 2 = e^{(t/τ)}

Now, taking ln on both sides,

\(\Rightarrow \ln 2 = \frac{{\rm{t}}}{{\rm{\tau }}}\)

Thus, the time constant is,

⇒ t = τ ln 2

**Calculation:**

Since, we know that in LC circuit, \({\rm{\tau }} = \frac{{\rm{L}}}{{\rm{R}}}\)

Given, Inductance L = 20 H,

Resistance, R = 10 Ω

On substituting the given values, we get,

\({\rm{\tau }} = \frac{{20}}{{10}} = 2\)

The time at which the rate of dissipation of energy (Joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor:

t = 2 ln 2