Two capacitors `4 mu F` and `6 mu F` in series are connected through a resistor of `10 k Omega` to a 18 V battery of negligible internal resistance. Ater a time of about 10 s, the battery is disconneted and capacitors are allowed to discharge through the resistance. Determine the voltage across each capacitor after a time lapse of 48 millisecond.

Explanation

Here, `C_(1) = 4 mu F, C_(2) = 6 mu F`,
`R = 10 k Omega = 10^(4) ohm, E = 18 V`.
As capacitors are connected in series
`(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) = (1)/(4) + (1)/(6) = (5)/(12)`
`C_(s) = (12)/(5) mu F = 2.4 xx 10^(-6) F`
Time constant of the circuit,
`tau = RC_(s) = 10^(4) xx 2.4 xx 10^(-6)`
`2.4 xx 10^(-2) s`.
As battery is connected for 10 seconds, which is very large comaperd to `tau` therefore, capacitors will acquire maximum voltage (18 V) and maximum charge `q_(0) = C_(s) E`
`= 2.4 xx 10^(-6) xx 18 = 4.32 xx 10^(-5) C`
During discharging, charge left is `
q = q_(0) e^(-t//tau)`
`= 4.32 xx 10^(-5) e^((-48 xx 10^(-3))/(2.4 xx 10^(-2)))`
`q = 5.832 xx 10^(-6) C`
As capacitor are in series, therefore, charge on each capacito is same (= q)
Potential difference/voltage across `C_(1)` is
`V_(1) = (q)/(C_(1)) = (5.832 xx 10^(-6))/(4 xx 10^(-6)) = 1.458 V`
Voltage acorss `C_(2)` is
`V_(2) = (q)/(C_(2)) = (5.832 xx 10^(-6))/(6 xx 10^(-6)) = 0.972V`