1cuz according to meh in d end V will bcum V/2
In the circuit shown the battery is ideal, with emf V. The capacitor is initially uncharged. The switch S is closed at time t=0.
We have to calculate the charge Q on the capapcitor at time t.
From my point of view, the cahrge Q should be given by,
Q = VC (1 - e -2t/3RC).
bcoz in case of charging , the charge at any time t is given by
Q=q max [ 1 - e - t/@ ] where @ is the time constant.
and q max is given by EC.
But the ans given at the back is,
Q = (VC /2)(1 - e -2t/3RC).
What's wrong with my concept??? Plzzz help. What's my mistake???
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9 Answers
1cuz 2 resistnces r in series so voltagge will b half.........
dnt kill meh if m wrng sorry
11But two resistances ( one across V and other near the negative terminal of V) aren in parallel and the equivalent of these is in series with the R (connected in series with C).
19@tushar.....you are still having the same problem.......you are still not answering but only questioning......anyways i should not interfere in what you do...but this is very unjustified !!! ....please don't mind !!!
1no while doin that u dnt consider d R in series wid C cuz currnt wnt flow thru that R
11At t = infinity,
the capactior acts as open circuit or a resistance of infinite value.
kk???
