Question

Q1

Five capacitors are connected across a potential difference Vab as shown in Figure. Because of the dielectrics used, each capacitor will break down if the potential across it is V=61 V. What should be the value of Vab (with one decimal place) if you reach the break down voltage one of the capacitor.

Q2
Suppose after charging a 3.7

Answer 1

The charge on a capacitor is given by Q = CV
Q is the charge, C is the capacitance and V the voltage.

Sine they are all in series they will all have the same charge.

For a given Q, the voltage will be highest across the smallest capacitor.

At break down, the 5uF capacitor will have a charge of 5 * 61 = 305 uC

The voltage on the 10 uF will be 305 / 10 = 30.5 v
on the 15 uF will be 305 / 15 = 20.33 v
on the 25 uF will be 305 / 25 = 12.2 v
and on the 45 uF will be 305 / 45 = 6.78 v

The total voltage Vab will be 61 + 30.5 + 20.33 + 12.2 + 6.78

= 130.81 volts.

Q2
V = (C1V1 + C2V2) / (C1 + C2)

We first need to find V2, volts across second capacitance:
C1 = 3.7, V1 = 5V, C2 = 12
Unconnected, V2 = 0
V = (3.7 * 5 + 12 * 0) / (3.7 + 12) = 1.942 volts
Take this charge and plug it in for V2

V = (3.7 * 5 + 12 * 1.942) / (3.7 + 12) = 2.662 V

Q3

let a=C1+Cx

V1 =Vo*C2/(C2+a)

1/(Vo/V1) =23.6*10^-6/(23.6*10^-6+a)

1/6 =23.6*10^-6/(23.6*10^-6+a)

23.6*10^-6+a=141.6*10^-6

a=118*10^-6

C1+Cx =118*10^-6

Cx =(118-3.5)*10^-6

Cx=11.45*10^-5 C