Question

4) Five capacitors are connected purely in parallel. Three of them have same amount of charge stored in them (all have Q), the fourth capacitor has 5.9 times more charge than Q, and the fifth capacitor has 3.6 times less charge than Q. If the equivalent capacitance of all the capacitors is 51 microfarads, what is the capacitance of the fifth capacitor in microfarads? 5) Five capacitors are connected purely in series to a 17 volt battery. Three of them have same amount of voltage across them of 3 Volts, the fourth capacitor is 2.7 times the capcitance any of the first three. If the equivalent capacitance of all five capacitors is 2.0 micro-farads, what is the stored energy in the fifth capacitor in micro-joules?

Answer 1

4)
Q1 = Q2 = Q3 = Q

Q4 = 5.9*Q

Q5 = Q/3.6

we know, charge across a capacitor, Q = C*V

and potential disfference across each capacitor is same in parallel connection.

Let C1 = C2 = C3 = C

C4 = 5.9*C

C5 = C/3.6

Cnet = C1 + C2 + C3 + C4 + C5

51 = C + C + C 5.9*C + C/3.6

==> C = 2.75 micro F

so, C5 = C/3.6

= 2.75/3.6

= 0.764 micro F

5)

V = 17 Volts

V1 = V2 = V3 = 3 V

let C1 = C2 = C3 = C

in series connection the charge on each capacitor is same.

so,

Q4 = Q1

C4*V4 = C1*V1

2.7*C1*V4 = C1*V1

V4 = V1/2.7

= 3/2.7

= 1.11 V

so, V5 = 17 - (3 + 3 + 3 + 1.11)

= 6.89 V

Q5 = Q1

C5*V5 = C1*V1

C5 = C1*V1/V5

= C*3/6.89

= 0.435*C

inseries combination,

1/Ceq = 1/C1 + 1/C2 + 1/C3 +1/C4 + 1/C5

1/2 = 1/C + 1/C + 1/C + 1/(2.7*C) + 1/(0.435*C)

==> C = 11.3 micro F

C5 = 0.435*C

= 0.435*11.3

= 4.91 micro F

energy stored in 5th capacitor = (1/2)*C5*V5^2

= (1/2)*4.91*10^-6*6.89^2

= 1.17*10^-4 J