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A capacitor C (=100 muF) is charged with...

 A capacitor `C (=100 muF)` is charged with a total charge of `1.2 xx 10^(-3)C` and connected in a circuit as shown. Find the current in the circuit after the key k is closed ( t= 0) as a function of time (t). Notice that the positively charged plate of the capacitor is connected to the negative terminal of the cell

Text Solution

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Let the current in the circuit be i(t), and the charge on the capacitor be q(t). Note that the initial polarity of charge `q_(0)` on the capacitor is opposite to the final steady state charge. Applying Kirchoff.s laws
`(q)/(C)+R((dq)/(dt))=E,`
`E = 12 V, R = 10 xx 10^(3)Omega C = 100 xx 10^(-6)F`
or,`((dq)/(dt))+(q)/(RC)=(E)/(R)`
Solving this, `q=C_(1) C_(1)e^(-t//RC)+CE` where is an arbitrary constant We determine C, from the initial condition, where `q_(0) = 1.2 xx 10^(-3)C`
`q = -q_(0) at t = 0, -q_(0) = C_(1) + CE`
Thus, `q(t)=-(CE+q_(o))e^(-1//RC)+CE`
The current`i=(dq)/(dt)=((CE+q_(0))/(RC))e^(-1//RC)`
Substituting, the values, `i = (2.4 mA)e^(-t//(sec)`
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