1) Calculate the limiting capacitance that may be joined directly across the 220V 50Hz mains supply without the risk of blowing a 13A fuse in the current. Does the capacitance represent the maximum or the minimum value?
2) A single component is connected across a 1.0kHz ac supply of r.m.s. voltage 12V. The r.m.s. current drawn from the supply is 0.2A. State the numerial quantity associated with the component if it is
a) a resistor
b) a capacitor
c) an inductor
2) F.6 Physics Electricity
2006-12-12 3:30 am
回答 (2)
2006-12-12 4:31 am
✔ 最佳答案
1. minimum valueXc = 1 / ωC
如果 C 增大，電抗 Xc 減小，電流會加大，fuse 會燒。
2. a) R = V / I = 12 / 0.2 = 60 Ω
b) Xc = 12 / 0.2 = 60 Ω = 1 / (2π1000C) ; C = 2.65μF
c) XL = 12 / 02 = 60 Ω = ωL ; L = 60 / (2π1000) ; L = 9.55 mH
2006-12-12 7:46 am
1)Minimum Reactance = 220 / 13 = 16.92 ohms
Using Xc = 1 / 2(pi)fC:
16.92 = 1 / 2(pi)(50)C
C = 188.1 uF
This C represents the maximum value of capacitance since if the value of C exceeds this vale, its reactance will decrease according to the equation and hence current will exceed 13A which will blow the fuse.
2)
a) Resistance value is independent of the frequence and hence the value can be directly obtained from Ohm's law:
V = IR
R = V/I
R = 12 / 0.2 = 60 ohms
b) Similar to a), the reactance value associated with the capacitor is 60 ohms and hence by the equation Xc = 1 / 2(pi)fC, C = 1 / 2(pi)(1000)(60) = 2.65 uF
c) Similar to a), the reactance value associated with the inductor is 60 ohms and hence by the equation XL = 2(pi)fL, L = 60 / 2(pi)(1000) = 9.55 mH
Using Xc = 1 / 2(pi)fC:
16.92 = 1 / 2(pi)(50)C
C = 188.1 uF
This C represents the maximum value of capacitance since if the value of C exceeds this vale, its reactance will decrease according to the equation and hence current will exceed 13A which will blow the fuse.
2)
a) Resistance value is independent of the frequence and hence the value can be directly obtained from Ohm's law:
V = IR
R = V/I
R = 12 / 0.2 = 60 ohms
b) Similar to a), the reactance value associated with the capacitor is 60 ohms and hence by the equation Xc = 1 / 2(pi)fC, C = 1 / 2(pi)(1000)(60) = 2.65 uF
c) Similar to a), the reactance value associated with the inductor is 60 ohms and hence by the equation XL = 2(pi)fL, L = 60 / 2(pi)(1000) = 9.55 mH
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