Answer:
Therefore the capacitance of the capacitor is 2.38 * 10⁻⁵F
Explanation:
We apply the concepts related to Power and energy stored in a capacitor.
By definition we know that power is represented as
[tex]P = \frac{E}{t}[/tex]
Where,
E= Energy
t = time
to find the Energy we have,
[tex]E = P*t[/tex]
[tex]P = 1*10^5Wt = 10*10^{-6}s[/tex]
[tex]E= (1*10^5)(10*10^{-6})E = 1.0J[/tex]
With the energy found we can know calculate the Capacitance in a capacitor through the energy for capacitor equation, that is
[tex]E=\frac{1}{2}CV^2[/tex]
[tex]C = \frac{2E}{V^2}\\\\\\\frac{2\times 1 }{290^2 - 3^2\\} \\= 2.38 \times 10^-^5F[/tex]
Therefore the capacitance of the capacitor is 2.38 * 10⁻⁵F
Answer:
24 μF.
Explanation:
Given:
Time, t = 10 μs
P = 1.0×105 W
V1 = 3 V
V2 = 290 V
Power, P = Energy/time
Energy = 1/2 × C × V^2
1 × 10-5 × 1 × 10^-5 = 1/2 × C × (290^2 - 3^2)
C = 2/84091
= 2.38 × 10^-5 F
= 24 μF