Three capacitors (C1 = 4.0µF, C2 = 3.0µF, and C3 = 2.0µF) are connected to a bat

Question

Three capacitors (C1 = 4.0µF, C2 = 3.0µF, and C3 = 2.0µF) are connected to a battery with E = 12 V and an internal resistance of r = 0.5 ?, as shown. All of the capacitors are initially uncharged.

A) How much time elapses after the switch is closed for the current through the battery to be 25% of its original value? (The original current is the current measured immediately after the switch is closed).

B) What is the terminal voltage of the battery when its current is 25% of its original value? (Again, the original current is the current measured immediately after the switch is closed).

C)  Once enough time has passed for all of the capacitors to be completely charged, how much energy is stored in the C3 = 2.0µF capacitor?

Explanation / Answer

A)

Take Io as the original current.
At time t, Io become 25% of Io,
According to the discharging equation of a capacitor,
25% Io = Io exp[-t/RC] ...(1)
C is the net capacitance in the circuit.

C2 and C3 are parallel, their resultant is C2 + C3 = 5µF.
C1 is in series with the above parallel combination,
The resultant capacitance, C = C1 x 5 / [C1+5] = 20/9 µF.

Solving equation (1) by putting the values of C as 20/9 µF, R = 0.5 ohm,
t = 1.386 RC
  = 1.386 x 0.5 x 20/9 x 10-6 = 1.54 x 10-6 s.

B)

Using the equation for the charging of capacitor,
Vc = Vo [ 1 - exp (-t/RC) ]
Vo = 12 V
t = 1.386 RC (time taken for the current to become 25% of its initial value)
Substituting,
Vc = 12 [ 1 - exp (-1.386) ]
   = 12 x ( 1 - 0.25) = 9 V.

C)

Charge in the capacitors, Q = VoC = 12 x 20/9 = 80/3 µC.
This charge is same for C1 and C2 + C3 combination.
Voltage across C2 + C3 = Q / C2 + C3 = (80/3) / 5 = 5.33 V
Energy stored in 2 µF capacitor = 0.5 C3 x 5.332
E = 0.5 x 2 x 10-6 x 5.332 = 28.44 x 10-6 J

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