A series ac circuit contains a 170-? resistor, a 11.0-mH...

A series ac circuit contains a 170-? resistor, a 11.0-mH inductor, a 4.10-?F capacitor, and an ac power source of voltage amplitude 45.0 V operating at an angular frequency of 360 rad/s. Part A What is the power factor of this circuit? cos ? = Part B Find the average power delivered to the entire circuit. P = Part C What is the average power delivered to the resistor, to the capacitor, and to the inductor? Enter your answers numerically separated by commas. PR, PC, PL =

Transcript

00:01 Hello everyone here the question consists three parts part a part b and part c so now coming to the first part of the question which is part a so in part a we have to find out the value of power factor of the circuit so here power factor is given by the formula p dot f equals to we can enter as cos 5 so further solving this from you can enter as cos multiplied by 10 inverse of x suffix l minus x x suffix c divided by r so further solving this form you can this we can enter power factor p .fx multiplied by 10 inverse of omega multiplied by l minus omega 1 divided by omega c whole thing divided by r so here we will put the value so therefore from here we will go the value of power factor so now here we will put the value so therefore we will get cos multiplied by 10 inverse of multiplied by the value of omega is 360 radian per second multiplied by the value of inductance l equals to given 11 multiplied by 10 to 0m .m.

01:13 Minus 1 divided by 360 multiplied by the value of capacitance c is 4 .1 multiplied by 10 2 .2 minus 6 ferr divided by the value of resistance r is 170 om so therefore further solving this from here we will go the value of power factor equals to 0 .245 so this is the answer for part -a so now coming to the next part of the question which is part b in part b of the question we need to find out the value of average power delivered to the entire circuit so the average power delivered is given by the formula p suffix average equals to i rms v rms multiplied by cost 5 so the rms current here it is given by the formula irms equals to and this is equation number 1, i divided by root 2.

02:22 So further solving this formula can't be the formula is v divided by z multiplied by root 2.

02:30 So further solving this from here we will go the value of irms current formula equals to g divided by square root of omega l minus 1 divided by omega c whole square plus r square multiplied by root 2 so therefore from here we will got rm s voltage vrrms equals to g divided by root 2 so in the equation 1 we will put this value so therefore we will got average power p a virus equals to 0 0 .357 1 so this is the answer for part b so now coming to the last part of the question which is part c in part c we have to find out the value of average power delivered to register capacitor and inductor so the average power delivered to the register you can return as p subcard equals to p equals to p .357 watt and average power delivered to capacitor can retain as p suffix c equals to 0 watt an average power delivered to inductor p suffix l equals to 0 1 so this is the answer for part 3 so therefore from here we can see that as our final answered for part a the value of power factor 5 equals to 0 .245 now coming to next part which is part b in part b the value of average power p suffix average equals to 0 .357 watt...

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