KVPY 2020-21 SB SX Exam Previous Papers

Question : 29 of 120

Marks: +1, -0

An initially uncharged capacitor C is being charged by a battery of emf E through a resistance R upto the instant when the capacitor is charged to the potential E/2, the ratio of the work done by the battery to the heat dissipated by the resistor is given by :-

2 : 1
3 : 1
4 : 3
4 : 1

Validate

Solution:

{i}={{E}}/{{R}} {e}^{-{t} \/ {RC}}, {Q}={CE}(1-{e}^{-{t} \/ {RC}})

Capacitor is charged to

{E}/{2}

,
So

Q={C E}/{2}

\∴{C E}/{2}=C E(1-{e}^{-t \/ R C})

{1}/{2}={e}^{-t \/ R C}

t={RC} l{n} 2

Work done by battery

=({Q}_{\text "flown "})(\Δ{V})

=({{CE}}/{2})({E})={{CE}^{2}}/{2}

Heat dissipated

=\∫↙{0} ↖{{RC} l{n} 2} {i}^{2} {Rdt}

={{E}^{2}}/{{R}} \∫↙{0} ↖{{RC} \ln 2} {e}^{-2 {t} \/ {RC}} \.{dt}

={3}/{4}({{CE}^{2}}/{2}\)

{\text "Work done "}/{\text" Heat dissipated "}={{CE}^{2} \/ 2}/{{3}/{4}({{CE}^{2}}/{2})}={4}/{3}

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