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A Source of 200 V - 50 Hz a.c. is connected to a resistance of 10 ohm, a capacitor C and an ammeter in series. If reading of ammeter is 2 A, what is the capacity of condenser ? |
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Answer» Here, `E_(v) = 200 V, v = 500 Hz, R = ohm, C = ? I = 2 A` From `Z = (E_(v))/(I_(v)) = (200)/(2) = 100 ohm` Again `R^(2) + X_(C)^(2) = Z^(2)` `X_(C) = sqrt(Z^(2) - R^(2)) = Z^(2)` `X_(C) = sqrt (Z^(2) - R^(2)) = sqrt(100^(2) - 10^(2)) = 99.5` As `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `:. C = (1)/(2 pi v X_(C)) = (1)/(2 xx 3.14 xx 50 xx 99.5) = (1)/(314 xx 99.5) = 3.2 xx 10^(-6) F = 3.2 mu F` |
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