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A parallel plate capacitor of plate separation `2 mm` is connected in an electric circuit having source voltage `400 V`. If the plate area is `60 cm^(2)`, then the value of displacement current for `10^(-6) sec` will beA. `1.062 amp`.B. `1.062xx10^(-2) amp`C. `1.062xx10^(-3) amp`D. `1.062xx10^(-4) amp` |
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Answer» Correct Answer - B `I_(D)=epsilon_(0)(delphi_(E))/(delt)=epsilon_(0)(EA)/(t)=(epsilon_(0)A)/(t)(v)/(d)` `I_(D)=(8.85xx10^(-12)xx400xx60xx10^(-4))/(2xx10^(-3)xx10^(-6))` `=1.062xx10^(-2)amp` Hence the corrent answer will be (`4`). |
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