Question

At the moment where we close the switch (t = 0), there is no charge on the
capacitor. C1 = 40μF and C2 = 60μF.
-diagram-
a) Find the intensity of the current in the battery when we close the switch.
b) Find the charge accumulated in capacitor C1 after 5 seconds.

Answer 1

Let these two capacitors are connected in series with 1 ohm resistance and 1 Volt battery.

So equivalent of the capacitors will be, $C = \frac{C_1C_2}{C_1+C_2}$ $= 24 \mu C$

Now,

Equation of the state,

$IR + \frac{Q}{C} = V \implies \frac{dQ}{dt} + \frac{Q}{RC} = \frac{V}{R}$

Solving above equation,

we get

$Q = VC + Ke^{-t/RC}$

applying condition that at t=0, Q = 0

then

$Q = VC(1 - e^{-t/RC} )$

(a) Current at t=0,

$I = \frac{dQ}{dt} = \frac{V}{R} = 1 A$

(b) Charge at time t= 5 s

$Q = VC(1 - e^{-t/RC} ) = 24\mu C$ (approximately)