IIT JEE , AIEEE Forum - Physics , Electricity

abhi_maks99

a capacitor of capacitance C is given a charge Q at time t=0. IT is then connected to a battery of emf E through a resistance R.Find charge on capacitor as a function of time???

abhi_maks99

ans. EC{1-e(raise to the power -t/RC) +Q{ e(raise to the power -t/RC)

Jyoti123

Is it an Irodov question?

abhi_maks99

no its ARIHANT

pink_ele

look '

since the capacitor is already charged and connected to EMF.

sp both charging and discarge will tace place

=charge left=charge gained+charge discharged

= EC{1-e(raise to the power -t/RC) +Q{ e(raise to the power -t/RC)

RITE!!!!!!!!!!!!!!!!

NO JYOTHI IT IS FRM HCV

iitkgp_bipin

At any time t the charge is given by :

q = A + Be^-t/T
where A and B are some constants to be found out and T is the time-constant of the circuit which is RC.

At t=0, q = Q
A + Be^-0/T = Q
A + B = Q ...................(1)

When t tends to infinity, potential difference across capacitor is E.
So q = EC
q = A + Be^-infinity/T = A = EC.............(2)

From (1) and (2) :
A = EC and B = Q - EC

Hence, q = A + Be^-t/T
q = EC + (Q-EC)e^-t/T

q = EC(1 - e^-t/RC) + Qe^-t/RC

Vinodh.rsp

q = A + Be^-t/T

How did this come?Plz Explain.

Vinodh

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