GATE Electrical Engineering (EE) 2019 Solved Paper

Question : 65 of 65

Marks: +1, -0

0.1 µ {F}

capacitor charged to

100 {V}

is discharged through a

1 {k} Ω

resistor. The time in

{ms}

(round off to two decimal places) required for the voltage across the capacitor to drop to

1 {V}

Solution:

V(0)=V_{0}=100 {V}

This is source free

1^{\;\text "st "\;}

order

R C

{v}( {t})= {V}_{0} {e}^{-0 {f}}

{V}_{0}=100 {V}

{t}= {R}_{ {eq}}- {C}=1000 × 0.1 × 10^{-6}

{v}( {t})=100 e^{\;{-t}/{π {/} {m}}}

{v}( {t})=100 {e}^{-10000 {t}}

1=100 {e}^{-10000 {t}}

{e}^{-10000 {t}}=\;{1}/{100}

-10000 {t}=\ln (\;{1}/{100})=-4.605

{t}=\;{4.605}/{10000}=0.46 {msec}

Go to Question:

Prev Question