(D) The potential difference across a discharging capacitor is given by $V = V_0 e^{-t/CR}$.Given $V_0 = 50 \, V$ and $V = 40 \, V$ at $t = 1 \, s$,we

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(D) The potential difference across a discharging capacitor is given by $V = V_0 e^{-t/CR}$.Given $V_0 = 50 \, V$ and $V = 40 \, V$ at $t = 1 \, s$,we have $40 = 50 e^{-1/CR}$,which implies $e^{-1/CR} = 4/5$.For option $(b)$,the potential difference after $t = 2 \, s$ is $V' = V_0 e^{-2/CR} = 50 (e^{-1/CR})^2 = 50 (4/5)^2 = 50 \times (16/25) = 32 \, V$. Thus,$(b)$ is correct.For option $(a)$,the energy stored in a capacitor is $U = \frac{1}{2} C V^2$. The fraction of energy remaining after $1 \, s$ is $U_f / U_i = (V_f / V_i)^2 = (40/50)^2 = (4/5)^2 = 16/25$. Thus,$(a)$ is correct.Since both $(a)$ and $(b)$ are correct,the correct option is $(d)$.

Class 12 Physics Electric Potential and Capacitance Charging and Discharging of Capacitance and RC circuit (DC)

Medium

$A$ parallel plate capacitor is charged to a potential difference of $50 \, V$. It is discharged through a resistance. After $1 \, s$,the potential difference between the plates becomes $40 \, V$. Then:

A
Fraction of stored energy after $1 \, s$ is $16/25$.
B
Potential difference between the plates after $2 \, s$ will be $32 \, V$.
C
Potential difference between the plates after $2 \, s$ will be $20 \, V$.
D
Both $(a)$ and $(b)$.

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Electric Potential and Capacitance

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Charging and Discharging of Capacitance and RC circuit (DC)

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$A$ parallel plate capacitor is charged to a potential difference of $50 \, V$. It is discharged through a resistance. After $1 \, s$,the potential difference between the plates becomes $40 \, V$. Then:

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In the given figure,the resistance of the coil of the galvanometer $G$ is $2\,\Omega$. The emf of the cell is $4\,V$. The ratio of the potential difference across $C_1$ and $C_2$ is:

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