NEET-XII-Physics
32: Electric Current in Conductors
- #8A 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 sdigAnsr: a,b,c,dAns : (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.
Disclaimer : Out of the four given options, two options are same.
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