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The voltage between the plates of a parallel plate capacitor of capacitance `1 muF` is changing at the rate of 8 V/s. What is the displacement current in the capacitor?A. `3 muA`B. ` 8 muA`C. `5 muA`D. `10 muA` |
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Answer» Correct Answer - B Displacement current `I_(D) = (epsi_(0)d_(phi0))/(dt)` but `phi = EA = (VA)/(d)` where A = area of the plates , d = distance between the plates `therefore I_(D) =epsi_(0) .(d)/(dt) ((VA)/(d)) = (epis_(0)A)/(d) (dv)/(dt)` but `c = (epsi_(0)A)/(d) `= capacity of the capacitor `therefore " " I_(D) = (Cdv)/(dt) = 10^(-6) xx 8` ` = 8 xx 10^(-6) A = 8 muA` |
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