Answer:
a) C = 4,012 10⁻¹⁴ F, b) Q = 1.6 10⁻¹¹ C
, c) U = 3.21 10⁻¹¹ J
Explanation:
a) The capacitance of a capacitor is
C = k e₀ A / d
Let's calculate
C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²
C = 4,012 10⁻¹⁴ F
b) let's look the charge
C = Q / ΔV
Q = C ΔV
Q = 4,012 10⁻¹⁴ 400
Q = 1.6 10⁻¹¹ C
c) The stored energy
U = ½ C ΔV²
U = ½ 4,012 10⁻¹⁴ 400²
U = 3.21 10⁻¹¹ J
Answer:
Your answer should be Incentives
Explanation:
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
Have everything in control and in order and discuss about different issues.
3.39 x 10^-13
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