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A parallel plate capacitor (figure) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s^(-1).a. What is the rms value of the conduction current ?b.Is the conduction current equal to the displacement current?c.Determine the amplitude of B at a point 3.0 cm from the axis between the plates. |
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Answer» Solution :`R=6.0 cm = 6xx10^(-3)m, C-100 pF = 100xx10^(-12)T` `V_(rms)=230V, omega = 300 rad s^(-1)` a.`X_(C )=(1)/(C omega)` `I_(rms)=(V_(rms))/(X_(C ))=V_(rms)xx C omega = 230xx100xx10^(-12)xx300=6.9xx10^(-6)A=6.9 MU A` B. Yes c.`B=(mu_(0)rI)/(2PI R^(2))`, But `I_(0)=sqrt(2)I_(rms)=sqrt(2)xx6.9xx10^(-6)=9.76xx10^(-6)A` `therefore B = (2xx10^(-7)xx3xx10^(-2)xx9.76xx10^(-6))/((6xx10^(-2))^(2))=1.63xx10^(-11)T` |
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