a) If you have five capacitors with capacitances 1.4 × 10 -6 F, 2.2 × 10 -6 F, 5
ID: 1405466 • Letter: A
Question
a) If you have five capacitors with capacitances 1.4 × 10-6 F, 2.2 × 10-6 F, 5.8 × 10-6 F, and two 6.8 × 10-6 F in series. What is the equivalent capacitance of all five?
C = _______F
b) 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 5.8 × 10-6 F capacitor?
Q = _______C
c)What is the potential difference between the plates of the 5.8 × 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 1.4 × 10-6 F, 2.2 × 10-6 F , 5.8 × 10-6 F, and two 6.8 × 10-6 F in parallel. What is the equivalent capacitance of all five?
C = _______F
f) If one attaches a 8 V battery to the system, how much charge is on the positive plate of the 5.8 × 10-6 F capacitor?
Q = _______C
g)What is the potential difference between the plates of the 5.8 × 10-6 F capacitor?
V = _______V
h) How much energy is stored in the entire capacitor system?
PE = _______J
Explanation / Answer
a)
C1 = 1.4 x 10-6 F , C2 = 2.2 x 10-6 , C3 = 5.8 x 10-6 , C4 = 6.8 x 10-6
net capacitance is given as
1/Ceq = 1/C1 + 1/C2 + 1/C3 + 1/C4 + 1/C4
1/Ceq = 1/(1.4 x 10-6 ) + 1/(2.2 x 10-6 ) + 1/(5.8 x 10-6 ) + 1/(6.8 x 10-6 ) + 1/(6.8 x 10-6 )
Ceq = 0.611 x 10-6 F
b)
In series charge remains same and is given as
Q = Ceq V = (0.611 x 10-6) (8) = 4.89 x 10-6 C
c)
V3 = Q/C3 = 4.89 x 10-6 / (5.8 x 10-6 ) = 0.843 volts
d)
E = (0.5) Ceq V2 = (0.5) (0.611 x 10-6) (8)2 = 19.6 x 10-6 J
e)
C1 = 1.4 x 10-6 F , C2 = 2.2 x 10-6 , C3 = 5.8 x 10-6 , C4 = 6.8 x 10-6
net capacitance in parallel is given as
Ceq = C1 + C2 + C3 + C4 + C4 = 1.4 x 10-6 + 2.2 x 10-6 + 5.8 x 10-6 + 6.8 x 10-6 + 6.8 x 10-6 = 23 x 10-6 F
f)
In parallel voltage remains same and charge stored in a capacitor is given as
Q3 = C3 V = (5.8 x 10-6) (8) = 46.4 x 10-6 C
g)
8 volts , same as the applied Voltage since capacitor and external Voltage are in parallel
h)
E = (0.5) Ceq V2 = (0.5) (23 x 10-6) (8)2 = 0.000736 J
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