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a,b, e are complete. Help on the others would be so appreciated!!

a) If you have five capacitors with capacitances 1.5 × 10-6 F, 3.3 × 10-6 F, 5.5 × 10-6 F, and two 6.2 × 10-6 F in series. What is the equivalent capacitance of all five?

C =F*
6.78e-7OK

b) Initially the capacitors are uncharged. Now a 14 V battery is attached to the system. How much charge is on the positive plate of the 5.5 × 10-6 F capacitor?

Q =C*
9.492e-6OK

c)What is the potential difference between the plates of the 5.5 × 10-6 F capacitor?

V =

d) How much energy is stored in the entire capacitor system?

PE =J

e) If you have five capacitors with capacitances 1.5 × 10-6 F, 3.3 × 10-6 F , 5.5 × 10-6 F, and two 6.2 × 10-6 F in parallel. What is the equivalent capacitance of all five?

C =F*
0.0000227OK

f) If one attaches a 14 V battery to the system, how much charge is on the positive plate of the 5.5 × 10-6 F capacitor?

Q =C

g)What is the potential difference between the plates of the 5.5 × 10-6 F capacitor?

V =V

h) How much energy is stored in the entire capacitor system?

PE =J

e) If you have five capacitors with capacitances 1.5 × 10-6 F, 3.3 × 10-6 F , 5.5 × 10-6 F, and two 6.2 × 10-6 F in parallel. What is the equivalent capacitance of all five?

C =F*
0.0000227OK

f) If one attaches a 14 V battery to the system, how much charge is on the positive plate of the 5.5 × 10-6 F capacitor?

Q =C

g)What is the potential difference between the plates of the 5.5 × 10-6 F capacitor?

V =V

h) How much energy is stored in the entire capacitor system?

PE =J

Respuesta :

Answer:

serie Ceq=0.678 10⁻⁶ F  and the charge Q = 9.49 10⁻⁶ C

Explanation:

Let's calculate all capacity values

a) The equivalent capacitance of series capacitors

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

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2

    1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2

    1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225

    1 / Ceq = 1,147

    Ceq = 0.678 10⁻⁶ F

b) Let's calculate the total system load

   Dv = Q / Ceq

   Q = DV Ceq

   Q = 14 0.678 10⁻⁶

   Q = 9.49 10⁻⁶ C

In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C

c) The potential difference

   ΔV = Q / C5

   ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶

   ΔV = 1,725 ​​V

d) The energy stores is

    U = ½ C V²

    U = ½ 0.678 10-6 14²

    U = 66.4 10⁻⁶ J

e) Parallel system

   Ceq = C1 + C2 + C3 + C4 + C5

   Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶

   Ceq = 22.7 10⁻⁶ F

f) In the parallel system the voltage is maintained

   Q5 = C5 V

   Q5 = 5.5 10⁻⁶ 14

   Q5 = 77 10⁻⁶ C

g) The voltage is constant V5 = 14 V

h) Energy stores

   U = ½ C V²

   U = ½ 22.7 10-6 14²

   U = 2.2 10⁻³ J

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