Question

An ultracapacitor is a very high-capacitance device capable of storing much more energy than an ordinary capacitor. It is designed so that the spacing of the plates is around 1000 times smaller than in ordinary capacitors. Furthermore the plates contain millions of microscopic nanotubes, which increase their effective area 100,000 times.

1) If the plate separation and effective area of a 10-μμF capacitor are changed as described above to make an ultracapacitor, what is its new capacitance? (ONE significant figure)

2) How much energy does the ultracapacitor store if charged to 1.5 V? (Express your answer to two significant figures.)

3) Compare this to the energy stored in an ordinary 10-μμF capacitor at that potential. (Express your answer to two significant figures.)

4)Compare the energy stored in the ultracapacitor to that of the AAA battery U = 6750 J. (Express your answer to two significant figures.)

5) If the ultracapacitor is to take 1.0 min to decrease to 1/e of its initial maximum charge, what resistance must it discharge through? (Express your answer to two significant figures.)

Answer 1

(1) We know that capacitance is given by
C = A /d
where A is plate area and d is seperation between them
Now Area is changed to = 100,000A and
distance is changed to = d/1000
C₁ = (100,000A)/(d/1000) = 10⁸*C = 10⁸*10*10^-6 = 1000 F
(2) Energy strored (E₁)= (1/2)C₁V² = (1/2)*1000*(1.5)² = 1125 J
(3) Energy in ordinary capacitor (E)= (1/2)CV² =(1/2)*(10*10^-6)*1.5² = 11.25*10^-6 J
E₁ /E = 10⁸
Hence 10⁸ times energy will be stroed in the ultracapacitor then ordinary capacitor .
(4) Energy stored in the battery (E₂) = 6750 J
E₁ /E₂ = 1125/6750 = (1/6)
E₁ = (1/6)E₂
hence energy stored in the ultracapacitor is (1/6) to that of battery.
(5)
Q = Q_o*(e^-t/RC)
(Q_o/e) = Q_o*(e^-t/RC)
e^-1 = e^-t/RC
1 = t/RC
RC = t = 60
r = 60 /(1000) = 0.06 ohm