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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 = 3 Omega. L = 25.48mH. And C = 796mu F. The impedance of the circuit |
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Answer» <P> 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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