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A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3Omega, L = 25.48 mH, and C = 796 muF. Find (a) the impedance of the circuit, (b) the phase difference between the voltage across the source and the current, (c) the power dissipated in the circuit, and (d) the power factor. |
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Answer» Solution :(a) To FIND the impedance of the circuit, we first calculate `X_(L) and X_(C )`. `X_(L)=2pi vL` `=2xx3.14xx50xx25.48xx10^(-3)Omega=8Omega` `X_(C )=(1)/(2pi vC)` `=(1)/(2xx3.14xx50xx796xx10^(-6))=4Omega` Therefore, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))=sqrt(3^(2)+(8-4)^(2))` `=5Omega` (b) Phase difference, `phi="tan"^(-1) (X_(C )-X_(L))/(R )` `=tan^(-1)((4-8)/(3))= -53.1^(@)` SINCE `phi` is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is `P=I^(2)R` Now, `I= (i_(m))/(sqrt(2))=(1)/(sqrt(2)) ((283)/(5))=40A` Therefore, `P=(40A)^(2)xx3Omega=4800W` (d) Power factor `=cosphi=cos(-53.1^(@))=0.6` |
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