Question

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor dissipates 2.0 j. what is the capacitance?

Answer 1

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:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$