1.

In the circuit diagram a capacitor which is initially uncharged is connected to an ideal cell of emf epsilon through a resistor 'R'. A leaky dielectric fills the space between the plates of dielectric. The capacitance of the capacitor with dielectric is C. Resistance of the dielectric is R' = R.

Answer»

Charge on the CAPACITOR as FUNCTION of time t is `(epsilon C)/(2)[1-e^(-(2t)/(RC))]`
Maximum charge on the capacitor is `(epsilon C)/(2)`.
When charge on the capacitor is maximum, then CURRENT in the circuit is `(epsilon)/(2R)`
All of the above options are true

Solution :(i) at `tgt0`

`I'`= current through dielectric
`=(q)/(C.R)` ….(i)
By `K.V.L. epsilon-iR-(q)/(c)=0` ..(2)
`i=I'+(dq)/(dt)=(q)/(RC)+(dq)/(dt)` ….(3)
By `(2)` and `epsilon-((q)/(RC)+(dq)/(dt))R-(q)/(c)=0`
`impliesepsilonC-2q-RC(dq)/(dt)=0`
`impliesepsilonC-2q=RC(dq)/(dt)impliesint_(0)^(q)(dq)/(epsilonC-2q)=int_(0)^(t)(dt)/(RC)`
`implies-(1)/(2)` ln `(epsilonC-2q)/(epsilonC)=(t)/(RC)impliesq=(epsilonC)/(2)(1-e^(2t)/(RC))`
(II) `q_(max)=(epsilonC)/(2)` as `trarroo`
and by (2) `epsilon-iR-(epsilon)/(2R)` at that time.


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