If you have five capacitors with capacitances 0.8×10 -6 F, 2.6×10 -6 F, 5.4×10 -
ID: 1344619 • Letter: I
Question
If you have five capacitors with capacitances 0.8×10-6 F, 2.6×10-6 F, 5.4×10-6 F, and two 9.8×10-6 F in series.
1)
What is the equivalent capacitance of all five?
C =
F
2)
Initially the capacitors are uncharged. Now a 8 V battery is attached to the system. How much charge is on the positive plate of the 5.4×10-6 F capacitor?
Q =
C
3)
What is the potential difference between the plates of the 5.4×10-6 F capacitor?
V =
V
4)
How much energy is stored in the entire capacitor system?
PE =
J
5)
If you have five capacitors with capacitances 0.8×10-6 F, 2.6×10-6 F, 5.4×10-6 F, and two 9.8×10-6 F in parallel. What is the equivalent capacitance of all five? What is the equivalent capacitance of all five?
C =
F
6)
If one attaches a 8 V battery is attached to the system. How much charge is on the positive plate of the 5.4×10-6 F capacitor?
Q =
C
7)
What is the potential difference between the plates of the 9.8×10-6 F capacitor?
V =
V
How much energy is stored in the entire capacitor system?
PE =
J
Explanation / Answer
(1)
the first three capacitor are in parallel so their equivalent is 0.8 muF + 2.6 muF + 5.4 muF = 8.8 muF
the two 9.8 muF capacitors are in series so their equivalent is 4.9 muF
now 4.9 muF and 8.8 muF are in series so their equivalent is 3.14 muF
(2)
V = 8 Volts
the charge on 5.4 muF will be
Q = C V
= (5.4 x 10-6) (8)
= 43.2 x 10-6 C
(3)
the potential difference across the plates is 8 V because it is in parallel to the 8 V battery
(4)
energy stored = (1 / 2) C V2
= (1 / 2) (3.14 x 10-6) (8)2
= 100.48 J
(5)
when two capacitors are in series their equivalent will be
Ceq = C1 C2 / (C1 + C2)
when two capacitors are in parallel their equivalent will be
Ceq = (C1 + C2)
use the same method that we have used in the options 1 to 4 to get the answers
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