In the given circuit the capacitor (C) m...

In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch `(S_1)`. Also when `(S_1)` is opend and `(S_2)` is closed the capacitor is connected in series with inductor (L).

At the start, the capicitor was uncharged. when switch `(S_1)` is closed and `(S_2)` is kept open, the time constant of this circuit is `tau`. which of the following is correct

After time interval `tau`, charge on the capacitor is `CV//2`

After time interval `2 tau`, charge on the capacitor of `CV(1 - e^(-2)]`

The work done by the voltage source will be half of the heat disspated when the capacitor is fully charged

After time interval `2 tau`, charge on the capacitor is `CV(1 - e^(-1)]`

Verified by Experts

The correct Answer is:

Charge on capacitor at time `t` is
`q = q_(0) (1 - e^(-t//tau))`
Here, `q_(0) = CV` and `t = 2 tau`
`q = CV(1 - e^(-2tau//tau)) = CV (1 - e^(-2))`

INDUCTANCE
CENGAGE PHYSICS|Exercise Archives (integer)|2 Videos
INDUCTANCE
CENGAGE PHYSICS|Exercise Subective Type|2 Videos
INDUCTANCE
CENGAGE PHYSICS|Exercise Archives (multiple Correct)|3 Videos
HEATING EFFECT OF CURRENT
CENGAGE PHYSICS|Exercise Thermal Power in Resistance Connected in Circuit|28 Videos
KINETIC THEORY
CENGAGE PHYSICS|Exercise Question Bank|31 Videos

In the given circuit the capacitor (C) may be charged through resistance R by a battery V by closing switch (S_1) . Also when (S_1) is opend and (S_2) is closed the capacitor is connected in series with inductor (L). When the capacitor gets charged compleely, (= (S_1) is opened and (S_2) is closed, Then,

at t=0 energy strored in the circuit is purely in the form of magnetic energy

at any time tgt0, current in the circuit is in the same direction

at tgt0, thre is no exchange of energy between the inductor and capacitor

at any time tgt0, instantaneous current in the circuit may be `Vsqrt((C)/(V))`

`Q=Q_(0)cos((pi)/(2)+(t)/(sqrt(LC)))`

`Q=Q_(0)cos((pi)/(2)-(t)/(sqrt(LC)))`

`Q=-LC(d^(2)Q)/(dt^(2))`

`Q=-(1)/(sqrt(LC))(d^(2)Q)/(dt^(2))`

after time interval `tau,` Charge on the capacitor is CV/2

after time interval `2 tau`, charge on the capacitor is CV`(1 - e^(-2))`

the work done by the voltae source will be half of the heat dissipated when the capacitor is fully charged.

after time interval `2 tau`, charge on the capacitor is CV`(1 - e^(-1))`