Question

Three capacitors C1 = 10.1 µF,

C2 = 23.0 µF,
and
C3 = 29.4 µF
are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Answer 1

energy is 0.14 J

Step-by-step explanation:

given data

capacitor C1 = 10.1 µF

capacitor C2 = 23.0 µF

capacitor C3 = 29.4 µF

charged = 125 V

to find out

maximum energy stored

solution

we will apply here formula that is

energy = 1/2 Q²/c ...............1

here c = 1 / ( 1/c1 + 1/c2 + 1/c3 )

c = 1 / ( 1/10.1 + 1/23 + 1/29.4 )

c = 5.66 µF

and Q = c1v1

Q = 10.1 × 125

Q = 1262.5 µC

so from equation 1

energy = 1/2 × 1262.5²/ 5.66= 140804.439

so energy is 0.14 J