Three capacitors C1 = 10.7 µF, C2 = 23.0 µF, and C3 = 29.3 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.

E = 1336.71875 J

Explanation:

We are given;. Capacitance of Capacitor 1; C1 = 10.7 µF

Capacitor 2; C2 = 23.0 µF

Capacitor 3; C3 = 29.3 µF

Supply voltage;V = 125 V

Formula for capacitance in series is;

Capacitors in series circuit: C(eq) = 1/C(1) +1/C(2) +1/C(3)

Thus, equivalent capacitance is;

C(eq) = (1/10.7) + (1/23) + (1/29.3) = 0.1711 µF = 0.1711 × 10^(6) F

Now, the formula for maximum energy stored is;

E = ½ × C(eq) × V²

E = ½ × 0.1711 × 10^(-6) × 125²

E = 1336.71875 J

answer: question1

1,000 tons

question2

3: 8

question 3

store d

step-by-step explanation:

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s