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Question
A capacitor of capacitance 500 μF is connected to a battery through a 10 kΩ resistor. The charge stored in the capacitor in the first 5 s is larger than the charge stored in the next.
(a) 5 s
(b) 50 s
(c) 500 s
(d) 500 s
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Solution
(a) 5 s
(b) 50 s
(c) 500 s
(d) 500 s
The charge (Q) on the capacitor at any instant t,
\[Q = CV(1 - e^{- t/RC} )\]
where
C = capacitance of the given capacitance
R = resistance of the resistor connected in series with the capacitor
RC = (10 × 103) × (500 × 10-6) = 5 s
The charge on the capacitor in the first 5 seconds,
\[Q_0 = CV(1 - e^{- 5/5} ) = CV \times 0 . 632 \]
The charge on the capacitor in the first 10 seconds,
\[Q_1 = CV(1 - e^{- 10/5} )\]
\[ Q_1 = CV(1 - e^{- 2} ) = 0 . 864 \times CV\]
Charge developed in the next 5 seconds,
Q' = Q1 - Q0
Q' = CV(0.864 - 0.632) = 0.232 CV
The charge on the capacitor in the first 55 seconds,
\[Q_2 = CV(1 - e^{- 55/5} )\]
\[ Q_2 = CV(1 - e^{- 11} ) = 0 . 99 \times CV\]
Charge developed in the next 50 seconds,
Q' = Q2 - Q0
Q' = CV(0.99 - 0.632) = 0.358 CV
Charge developed in the first 505 seconds,
\[Q_3 = CV(1 - e^{- 500/5} ) = CV(1 - e^{- 100} ) \approx CV\]
Charge developed in the next 500 seconds,
Q' = CV (1 - 0.632) = 0.368 CV
Thus, the charge developed on the capacitor in the first 5 seconds is greater than the charge developed in the next 5,50, 500 seconds.
Notes
Out of the four given options, two options are same.
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