(a) Capacitive resistance $$X_c = dfrac{1}{wC} = dfrac{1}{2 pi vC} = dfrac{1}{2 pi imes 50 imes dfrac{100}{ pi} imes 10^{-6} } = 100Omega$$  Impedance of the circuit , $$Z = sqrt{R^2 + X^2 _C} = sqrt{100^2 + 100^2} = 100sqrt2$$ Current in the circuit $$I_{rms} = dfrac{E_{rms}}{Z} = dfrac{220}{100sqrt2} = 1.56 A$$ (b) Voltage across resistor $$V_R = I_{rms} R = 1.56 imes 100 = 156 V$$      Voltage across capacitor, $$V_C = I_{rms} imes C = 1.56 imes 100 = 156 V$$ The algebric sum of voltages across the combination is                             $$V_{rms} = V_R + V_c = 156 + 156 = 312 V$$ While $$V_{rms}$$ of the source is $$220V$$ . Yes the voltage across the combination is more than the voltage of the source. The voltage across the resistor and the capacitor are not in phase. This paradox can be resolved as when the current passes through the capacitor , it lead the voltage $$V$$ , phase $$dfrac{pi}{2}$$. So, voltage of the source can be given as                        $$V_{rms} $$=$$ sqrt{{V_R}^2 + {V_C}^2} $$=$$ sqrt{(156)^2 + (156)^2} = 156sqrt2 = 220V$$