1. [tex]4.1\cdot 10^7 J[/tex]
The energy stored in a capacitor is given by
[tex]U=\frac{1}{2}CV^2[/tex]
where
C is the capacitance
V is the potential difference across the capacitor
In this problem, we have
[tex]C=32 mF = 0.032 F[/tex] is the capacitance of the capacitor
[tex]V=16 kV=16000 V[/tex] is the p.d. across the capacitor
Solving the equation, we find the energy stored:
[tex]U=\frac{1}{2}(0.032 F)(16000 V)^2=4.1 \cdot 10^7 J[/tex]
2. [tex]4.1\cdot 10^9 W[/tex]
The average power delivered to the coils is equal to the ratio between the energy released and the time taken:
[tex]P=\frac{E}{t}[/tex]
where
[tex]E=4.1 \cdot 10^7 J[/tex] is the energy released by the capacitor
[tex]t = 10 ms = 0.01 s[/tex] is the time taken
Substituting into the formula, we find
[tex]P=\frac{4.1\cdot 10^7 J}{0.01 s}=4.1\cdot 10^9 W[/tex]
A) The energy stored in the capacitor bank when fully charged = 4.1 * 10⁷J
B) The average power delivered to the coils during this discharge = 4.1 * 10⁹ Watts
Given data :
Total capacitance of capacitor bank = 32 mF
Charge on capacitors = 16 kV
applying the equation below
U = [tex]\frac{1}{2} CV^{2}[/tex] --- ( 1 )
where : C = 32 mF, V = 16000 V,
insert values into equation ( 1 )
U = 1/2 * ( 0.032) F * ( 16000 )²
= 4.1 * 10⁷J
applying the equation below
Power ( P ) = [tex]\frac{E}{T}[/tex] ----- ( 2 )
where : E = 4.1 * 10⁷J , T = 0.01 s
therefore :
P = 4.1 * 10⁷ / 0.01
= 4.1 * 10⁹ Watts
Hence we can conclude that A) The energy stored in the capacitor bank when fully charged = 4.1 * 10⁷J and The average power delivered to the coils during this discharge = 4.1 * 10⁹ Watts
Learn more about energy storage in capacitors : https://brainly.com/question/9434451
