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A resistor of `200 Omega`and a capacitor of `15.0muF` are connected in series to a `220V`,` 50Hz` source. (a) Calculate the current in the circuit . (b) Calcutalte the voltage (rms) across the resistor and the inductor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. |
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Answer» Here, `R = 200 ohm`, `C = 15.0 mu F = 15 xx 10^(-6) F` `E_(v) = 220 V, v = 50 Hz, I_(v) = ?` `V_(R ) = ?, V_(C ) = ?` Now `X_(C ) = (1)/(omega C) = (1)/(2 pi v C)` `= (1)/(2 xx 3.14 xx 50 xx 15 xx 10^(-6)) = 212.3 Omega` (a) Impedance of the circuit, `Z = sqrt(R^(2) + X_(C)^(2))` `= sqrt(2000^(2) + (212.3)^(2))` `= 291.7 ohm` `:.` Current in the circuit, `I_(v) = (E_(v))/(Z) = (220)/(291.7) = 0.75 A` (b) `V_(R ) = I_(v) xx R = 0.75 xx 200 = 150.8 V` `V_(C ) - I_(v) X_(C ) = 0.75 xx 212.3 = 159.2 V` `V_(R ) + V_(C ) = 150.8 + 159.2 = 310 V`, which is more than the source voltage of 220 V. This paradox is resolved by the fact that the two voltages are not in same phase. Therefore, they connot be added like ordinary number. As `V_(R )` and `V_(C )` are out of phase by `90^(@)`, therefore, `V_(RC) = sqrt(V_(R )^(2) + V_(C )^(2)) = sqrt((150.8)^(2) + (159.2)^(2))` `= 220 V`, the source voltage. |
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