Suppose you are trying to design a capacitor-based energy-storage device. You wo

Question

Suppose you are trying to design a capacitor-based energy-storage device. You would like to be able charge the capacitor from a USB port of your computer (the USB voltage is 5.0 Volts) and you plan to use a super-capacitor of 1.5F to store energy. (A super-capacitor is a real term. Google it if you are interested in knowing what it is and how it works. For the purpose of this problem, a super-capacitor is just a large capacitor. ) 1. What resistor would you use in the charging circuit (shown on the left) so that the current does not exceed 1.0 A (as you don

Explanation / Answer

1)

Initially when you switch on the circuit,

the capacitor acts as a short circuit, so the circuit will act as if there is no capacitora and there is only a resistor,

So, let the resistance of the circuit be R

So, current in the circuit = E/R

here E = emf = USB voltage = 5 V

So, for a current of 1.0 A which is the maximum current when the circuit is just turned on.

So, 1 = 5/R

So, R = 5 = 5 ohm <-----------answer

2)

while charging, the charge on the capacitor is given by :

Q = Q0*(1-e^(-t/RC)) <------------- (1)

where Q0 = final charge after a long time which remains constant while the circuit is remained turned on which is the maximum charge

t= time

R = resistance = 5 ohm

C= capacitance = 1.5 F

We know, the energy stored in capacitor,

U = 0.5*Q^2/(2*C)

So, maximum energy will be stored when there is maximum charge(Q0) (as C(capacitance) is constant)

So, maximum energy, Umax = 0.5*(Q0)^2/(2*C)

So, for 99 percent of energy , U = 99 percent of Umax

So, U = (99/100)*(0.5*(Q0)^2/(2*C)) = 0.5*Q^2/(2*C)

So, Q = sqrt(0.99)*Q0

So, it means when charge is sqrt(0.99)*Q0 , the energy is 99 percent of maximum

So, lets put it in equation (1), to calulate the time taken

So, sqrt(0.99)*Q0 = Q0*(1-e^2(-t/RC))

So, sqrt(0.99) = (1-e^(-t/(5*1.5)))

So, t = 39.72 s <--------------answer

3)

for discharge condition,

the charge stored is given by,

Q = Q0*(e^(-t/RC))

here R = total resistance = (1.3*10^6 + 0.2) ohm

We know, energy stored, U = Q^2/(2*C)

So, U = (Q0*(e^(-t/RC))^2/(2*C)) = Q0^2*(e^(-2t/RC))/(2*C) -----------(2)

Also, for initial charge(Q0), initial energy,U0 = Q0^2/(2*C)

so, for 50 percent of initial energy,

U50 = (50/100)*(U0) = 0.5*(Q0^2/(2*C))

puttin this in equation (2),we get

0.5*(Q0^2/(2*C)) = Q0^2*(e^(-2t/RC))/(2*C)

So, 0.5 = e^(-2t/RC)

so, 0.5 = e^(-2*t/((1.3*10^6+5)*(1.5)))

So, t = 6.76*10^5 seconds = 6.76*10^5/(3600) = 187.8 hours

So, we have to charge the capacitor in every 187.8 hours <--------------answer

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