Capacitors in Series problems and solutions

 Problem #1

Three capacitors C₁ = 100μF, C₂ = 220 μF and C₃ = 470 μF connected with 20 V batteries. Determine (a) capacitor total capacity, (b) charge and potential difference of each capacitor, and (c) total charge!

Answer;
(a) The total capacity for the series of capacitors arranged in series is

1/C_total = 1/C₁ + 1/C₂ + 1/C₃

1/C_total = 1/(100 μF) + 1/(220μF) + 1/(470μF) = 1724/(103400 μF)

C_total = 60 μF

(b) Because it is arranged in series, the charge for each capacitor is the same which is equal to the source charge

q₁ = q₂ = q₃ = C_totalV = 60 μF x 20 Volt = 1200 μC

The potential difference of each capacitor is calculated by the equation V = q/C,

V₁ = q/C₁ = 1200μC/(100μF) = 12V

V₂ = q/C₂ = 1200μC/(220μF) = 5.45 V

V₃ = q/C₃ = 1200μC/(470μF) = 2.55 V

Problem #2
In the capacitor circuit below C₁ = 4 μF, C₂ = 6 μF, C₃ = 12 μF, and C₄ = 2 μF. Field 1 is given a charge of 400 μC, field VIII is grounded, and the distance between 2 pieces of capacitors is 2 mm, 2 mm, 4 mm and 8 mm, respectively. Calculate: (a) Potential of each chip and (b) The strength of the electric field between the pieces of the capacitor!
 
Answer:
What needs to be considered in this problem is that

• The earthed chip is potentially zero (the earth we take as a reference)
• The chips that are connected with wire have the same potential
• All pieces are equal to q = 400 μC because they are connected in series

First we look for the total potential (the difference in voltage between chip I and chip VIII), by dividing the total load of the chip by the chip capacitor.

Total chip capacity is

1/C_Total = 1/C₁ + 1/C₂ + 1/C₃ =  ¼ + 1/6 + 1/12 = 1

C_Total = 1 μF

Then,

V_total = Q_total/C_total = 400 x 10^-6/10^-6 = 400 Volt

V_total = V_I - V_VIII

400 = V_I - 0

V_I = 400 volts

V₁ = Q₁/V₁ = 400 x 10^-6/(4 x 10^-6) = 100 volt

V₁ = V_I - V_II

100 = 400 - V_II

V_II = 300 volt

V_III = V_II = 300 volt (CONNECTED)

V₂ = Q₂/C₂ = 400 x 10^-6/(6 x 10^-6) = 200/3 volts

V₂ = V_III - V_IV

200/3 = 300 - V_IV

V_IV = 700/3 volt

V_V = V_IV = 700/3 volts (CONNECTED)

V₃ = Q₃/C₃ = 400 x 10^-6/(12 x 10^-6) = 100/3 volt

V₃ = V_V - V_VI

100/3 = 700/3 - V_VI

V_VI = 200 volt

V_VII = V_VI = 200 volts (connected)

V₄ = Q₄/C₄ = 400 x 10^-6/(2 x 10^-6) = 200 volt

(b) To calculate the electric field of a capacitor, we use the formula E = V/d.

E₁ = V₁/d₁ = 100/(2 x 10^-3) = 5 x 10⁴ N/C

E₂ = V₂/d₂ = (200/3)/(2 x 10^-3) = 3.33 x 10⁴ N/C

E₃ = V₃/d₃ = (100/3)/(4 x 10^-3) = 8.3 x 10⁴ N/C

E₄ = V₄/d₄ = 200/(8 x 10^-3) = 2.5 x 10⁴ N/C   

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