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A circuit containing a 80 mH inductor and a `60 mu F` capacitor in series is connected to a 230 V 50 Hz supply. The resistance of the circuit is negligible. Obtain the rms values of protential drops across each element. |
Answer» Given Inductance `L = 80 mH = 80xx10^(-3)H` Capacitance of capacitor `C = 60 mu F = 60xx10^(-6)F` The rms value of voltage `V_("rms")=230 V` Frequency f = 50 Hz , Resistance R = 0 Potential drop across L `V_("rms")L=I_("rms")xx X_(L)=8.29xx2xx3.14xx50xx80xx10^(-3)` `= 208.25 V` Potential drop across C `V_("rms")C=I_("rms")xx X_(C )` `= 8.29xx(1)/(2xx3.14xx50xx60xx10^(-6))` `= 440.02 V` `therefore` Applied rms voltage `= V_(C )-V_(L)` `= 440-208.25=231.75 V` |
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