) If you have five capacitors with capacitances 1.3 × 10 -6 F, 3.4 × 10 -6 F, 5.
ID: 2001116 • Letter: #
Question
) If you have five capacitors with capacitances 1.3 × 10-6 F, 3.4 × 10-6 F, 5.1 × 10-6 F, and two 6.3 × 10-6 F in series. What is the equivalent capacitance of all five?
C = F
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 5.1 × 10-6 F capacitor?
Q = C
c)What is the potential difference between the plates of the 5.1 × 10-6 F capacitor?
V = V
d) How much energy is stored in the entire capacitor system?
PE = J
e) If you have five capacitors with capacitances 1.3 × 10-6 F, 3.4 × 10-6 F , 5.1 × 10-6 F, and two 6.3 × 10-6 F in parallel. What is the equivalent capacitance of all five?
C = F
f) If one attaches a 14 V battery to the system, how much charge is on the positive plate of the 5.1 × 10-6 F capacitor?
Q = C
g)What is the potential difference between the plates of the 5.1 × 10-6 F capacitor?
V = V
h) How much energy is stored in the entire capacitor system?
Explanation / Answer
We use the expressions for capacitors
. a) 1/Ceq = 1/C1+ 1/C2+1/C3+1/C4+1/C5 1/Ceq = 1/1.3+1/3.4+/5.1+1/6.3+1/6.3
1/Ceq= (3.4 5.1 6.3 + 1.3 5.1 6.3 + 1.3 3.4 6.3 +1.3 3.4 5.1 + 1.3 3.4 5.1)/( 1.3 3.4 5.1 6.3)
1/ Ceq= (109.242+ 41.769+ 27.846+ 22.542+ 22.542)/(142.0146 10-6)= 223.941/ 142.0146 10-6
Ceq= 142.0146 10-6 / 223.941
Ceq= 0.634 10-6 F
. b) V= 14 V
Qt = Ceq V Qt = 0.634 10-6 14 Qt = 8.876 10-6 C
In a circuit of series capacitors is charged all capacitors equals Q5 = Qt
Q5 = 8.876 10-6 C
. c) V5 = Q/C5 V5= 8.876 10-6 C / 5.1 10-6 F V5 =1.74 V
. d) energy
U = ½ Ceq V2 U = ½ 0.634 10-6 142 U= 62.132 10-6 J
. e) parallel capacitors
Ceq = C1+C2+C3+C4+C5 Ceq= (1.3 +3.4+5.1+6.3+6.3) 10-6 F Ceq= 22.4 10-6 F
. f) V= 14 V Q5 = C5 V
In parallel the voltages applied to all capacitors are equal
Q5 = 5.1 10-6 14 Q5= 71.4 10-6 C
.g) In a parallel circuit voltages they are equal V5=V = 14V
.h) U = ½ Ceq V2 U= ½ 22.4 10-6 142 U= 2195.2 10-6 J
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