Answer:
The capacite is C=5.32 uF using the equations of voltage and energy in capacitance
Explanation:
The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J
Using:
[tex]V_{t}= V_{o}e^{\frac{-t}{R*C} }[/tex]
Voltage in this case is the energy dissipated so
[tex]E_{t}= E_{o}e^{\frac{-t}{R*C} }[/tex]
[tex]\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }[/tex]
[tex]\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }[/tex]
Using the equation to find capacitance
[tex]ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }[/tex]
[tex]C= 5.32x10^{-6}[/tex] F
C= 5.32 uF because u is the symbol for micro that is equal to [tex]10^{-6}[/tex]
