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An alternating voltage `E=200sqrt(2) sin (100 t)V` is connected to a `1 (mu)F` capaacitor through an ac ammeter (it reads rms value). What will be the reading of he ammeter?A. `10 mA`B. `5 mA`C. `5 sqrt(2)mA`D. `10 sqrt(2)mA` |
Answer» Correct Answer - B `X_(C) = 1/(omega C) =(1)/(100 xx 10 ^(-6)) = 10^(4) Omega` so, ` I_(rms) = (E_(rms))/(Z) = (50 sqrt(2))/(sqrt(2) xx 10^(4)) = 5 mA`. |
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