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

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Step-by-step explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

`V_(t)= V_(o)e^{(-t)/(R*C) }`

Voltage in this case is the energy dissipated so

`E_(t)= E_(o)e^{(-t)/(R*C) }`

`(√(E_t) )/(√(E_o) ) = e^{(-t)/(R*C) }`

`(√(3 J) )/(√(5J) ) = e^{(-13.6ms)/(10kw*C) }`

Using the equation to find capacitance

`ln 0.775= e^{(-13.6 x10^(3) )/(10x10^(3)*C )} \\ln(0.775)= ln * e^{(-13.6 x10^(3s) )/(10x10^(3)*C )} \\\\ln(0.775)= {(-13.6 x10^(3) )/(10x10^(3)*C )} \\C= (-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)`