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A parallel plate capacitor is charged to a potential difference of $50\, V$. It is discharged through a resistance. After $1$ second, the potential difference between plates becomes $40 \,V$. Then
Fraction of stored energy after $1$ second is ${16}/{25}$
Potential difference between the plates after $2$ seconds will be $32\, V$
Potential difference between the plates after $2$ seconds will be $20\, V$
Both $(a)$ and $(b)$
Solution
(d) By using
$V = {V_0}{e^{ – t/CR}} \Rightarrow 40 = 50\,{e^{ – 1/CR}} \Rightarrow {e^{ – 1/CR}} = 4/5$
Potential difference after $2$ $sec$
$V' = {V_0}{e^{ – 2/CR}} = 50{({e^{ – 1/CR}})^2} = 50{\left( {\frac{4}{5}} \right)^2} = 32\,V$
Fraction of energy after $1$ $sec$
$ = \,$$\frac{{\frac{1}{2}C{{({V_f})}^2}}}{{\frac{1}{2}C{{({V_i})}^2}}} = {\left( {\frac{{40}}{{50}}} \right)^2} = \frac{{16}}{{25}}$
