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                                    A 10 mu F capacitor is in series with a 50 Omega resistance and the combination is connected to a 220 V, 50 Hz line. Calculate (i) the capacitivereactance, (ii) the impedance of the circuit and (iii) the current in the circuit. | 
                            
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Answer» Solution :Here, `C = 10 mu F = 10 xx 10^(-6) = 10^(-5)F`  (i) Capacitive reactance, `X_(C) = (1)/(omega C) = (1)/(2pi v C) = (1)/(2 xx 3.14 xx 50 xx 10^(-5)) = 318.5 Omega` (ii) Impedance of CR CIRCUIT. `Z_(CR) = sqrt(R^(2) + X_(C)^(2)) = sqrt((50)^(2) + (318.5)^(2))` `= 322.4 Omega` (III) Current, `I_(rms) = (E_(rms))/(Z_(CR)) = (220)/(322.4) = 0.68 A`  | 
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