If you have five capacitors with capacitances 1.3×10 -6 F, 1.7×10 -6 F, 4.3×10 -
ID: 581723 • Letter: I
Question
If you have five capacitors with capacitances 1.3×10-6 F, 1.7×10-6 F, 4.3×10-6 F, and two 8.2×10-6 F in series.
1)
What is the equivalent capacitance of all five?
C =
2)
Initially the capacitors are uncharged. Now a 13 V battery is attached to the system. How much charge is on the positive plate of the 4.3×10-6 F capacitor?
Q =
3)
What is the potential difference between the plates of the 4.3×10-6 F capacitor?
V =
4)
How much energy is stored in the entire capacitor system?
PE =
5)
If you have five capacitors with capacitances 1.3×10-6 F, 1.7×10-6 F, 4.3×10-6 F, and two 8.2×10-6 F in parallel. What is the equivalent capacitance of all five? What is the equivalent capacitance of all five?
C =
6)
If one attaches a 13 V battery is attached to the system. How much charge is on the positive plate of the 4.3×10-6 F capacitor?
Q =
7)
What is the potential difference between the plates of the 8.2×10-6 F capacitor?
V =
8)
How much energy is stored in the entire capacitor system?
PE =
Explanation / Answer
1.
combination of capacitances in series is given as
1/Ctotal = 1/C1 + 1/C2 + 1/C3 + 1/C4 + 1/C5
1/Ctotal = 1/1.3 + 1/1.7 + 1/4.3 + 1/8.2 + 1/8.2
Ctotal = 0.55 uF
2.
V = battery Voltage = 13
Net charge coming from the battery is given as
Q = Ctotal V = 0.55 x 13 = 7.15 uC
Since same charge flows in series , so charge = 7.15 x 10-6 C
3.
potential differenc = charge / Capacitance = 7.15 x 10-6 / (4.3 x 10-6) = 1.66 volts
4.
Total energy = (0.5) Ctotal V2 = (0.5) (0.55) (13)2 = 46.5 J
5.
combination of capacitances in series is given as
Ctotal = C1 + C2 + C3 + C4 + C5
Ctotal = 1.3 + 1.7 + 4.3 + 8.2 + 8.2 = 23.7 uF
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