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When an alternating voltage of 220 V is applied across a device X, a current of 0.25 A flows which lags behind the voltage in phase by pi/2 rad. If the same voltage is applied across another device Y, the same current flows but now it is in phase with the applied voltage. (a) Name the devices X and Y. (b) Calculate the current flowing in the circuit when the same voltage is applied across the series combination of X and Y. |
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Answer» SOLUTION :(a) As in device X the current I lags behind voltage V by `pi/2`the device X is an inductor L. In device Y the current I and voltage V are in phase, the device Y is a resistor. (b) `X_(L) = R = (220 V)/(0.25 A) = 880 Omega` When L and R both are connected in series, the total impedance, `Z = sqrt(R^(2) + X_(L)^(2))` `therefore` Current `I = V/Z = V/sqrt(R^(2)+ X_(L)^(2)) = 220/sqrt((800)^(2) + (880)^(2)) = 220/(800sqrt(2)) = 1/(4sqrt(2))A = 0.18 A` |
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