GATE Electronics and Communications (EC) 2019 Solved Paper

Question : 60 of 65

Marks: +1, -0

The RC circuit shown below has a variable resistance

{R}( {t})

given by the following expression:

R(t)=R_{0}(1-\;{t}/{T}) \;\text " for "\; 0 ≤ {t}< {T}

where

{R}_{0}=1 Ω

, and

{C}=1 {F} .

We are also given that

{T}=3 {R}_{0} {C}

and the source voltage is

{V}_{ {s}}=1 {V} .

If the current at time

{t}=0

1 {A}

, then the current

{I}( {t})

, in amperes, at time

{t}= {T} {/} 2

is _________ (rounded off to 2 decimal places).

Solution:

We know that,

{I}_{ {C}}( {t})= {C} \;{ {d} {V}_{ {C}}( {t})}/{ {dt}}

And capacitor will be charged by the following equation

V_{c}(t)=V_{s}(1-e^{-t {/} τ})

I_{C}(t)=C ⋅ \;{d}/{d t}[V_{S}(1-e^{-t {/} τ})]

∴ I_{C}(t)=\;{V_{S}}/{R(t)} e^{-t {/}R(t) ⋅ C}

Given

{R}( {t})= {R}_{\o}[1-\;{ {t}}/{ {T}}]

Now

{R}_{\o}=1

and

{C}=1

∴ T=3 R_{\o} C=3

∴ {R}( {t})=[1-\;{ {t}}/{3}]

\& I_{C}(t)=\;{1}/{(1-\;{t}/{3})} × e^{\;{-t}/{(1-\;{t}/{3})}}

t=\;{T}/{2}=\;{3}/{2} \sec

{I}_{ {C}}( {t})=2 e^{-3}

=0.099

{I}_{ {C}}( {t}) ≈ 0.1 {mA}

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