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An electrical device draws 2 kW power form AC mains [voltage 223 V (rms) `= sqrt(50,000) V`]. The current differs (lags) in phase by `phi (tan = (-3)/(4))` as compared to voltage. Find (i) R, (ii) `X_(C ) - X_(L)`, and (iii) `I_(M)`. Another device has twice the values for R, `X_(C )` and `X_(L)`. How are the answers affected ? |
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Answer» Here, `P = 2 kW = 2000 W. E_(v) = sqrt(50000) = 223 V, tan phi = - (3)/(4) , R = ? , X_(C ) - X_(L) = ?, I_(M) = I_(0) = ?` From `P = (E_(upsilon)^(2))/(Z) , Z = (E_(upsilon)^(2))/(P) = (50000)/(2000) = 25 Omega` Now, `tan phi = (X_(C ) - X_(L))/(R ) = - (3)/(4) :. X_(C ) - X_(L) = - (3)/(4) R` From `Z^(2) = R^(2) + (X_(C ) - X_(L))^(2)` `(25)^(2) = R^(2) + (-(3)/(4) R)^(2) = (25)/(16) R^(2)` `R = (4)/(5) xx (25) = 20 ohm` From (i), `X_(C ) - X_(L) = - (3)/(4) xx 20 = - 15 Omega` Now, `I_(upsilon) = (E_(upsilon))/(Z) = (223)/(25) ~= 9 A` `I_(0) = sqrt(2) I_(upsilon) = sqrt2 xx 9 A = 12.6 A` When R, `X_(C ), X_(L)` are all doubled, `tan phi` does not change , Z is doubled, current is halved and power drawn is halved. |
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