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At=0 the switch is closed. What is time constant. Find Qmax after the closing of the switch and find the current in the resistor after 10 seconds

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As at t=0 switch is open so initially capacitor is uncharged.So,we use equation of charging of capacitor in an RC Circuit which is:

$Q(t)=Q_0[1-e^{t/T}]$

here $Q_0$ =CE where c is the capacitance=$6 \mu f$ and e=battery potential =24 v (according to given question)

and T=time constant =RC i.e. product of resistance and capacitance = 54micro seconds (according to given question)

so maximum charge=$Q_0$=144 micro coulomb.

current in the circuit after t seconds is given by following equation

$I(t) = (Q_o/RC) e^{-t/T}$

=2.66*e^(-10/.000054)=0A as the capacitor is fully charged.

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