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An $R L C$ series circuit has a 1.00 $\mathrm{k} \Omega$ resistor, a 150$\mu \mathrm{H}$ inductor, and a 25.0 $\mathrm{nF}$ capacitor. (a) Find the power factor at $f=7.50 \mathrm{Hz}$ . (b) What is the phase angle at this frequency? (c) What is the average power at this frequency? (d) Find the average power at the circuit's resonant frequency.

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Physics 102 Electricity and Magnetism

Chapter 23

Electromagnetic Induction, AC Circuits, and Electrical Technologies

Electromagnetic Induction

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Problem 104 pi in this r l c series circuit we have been given that is equal to 1 kilmindectance is given as 150 macrohenryand keester, that is in it 25 men of rode, so we have found in part a power factor of the circuit. That is given that frequency of the applied voltage is 7.5 hurtandas. We know that power factor is equal to, or was that that is resistance in the circuit or the impudence of the circuit. So we have to find impudence first for that we have formula equal to. I square plus x, l minus x whole square, so we will find first excellent x. We have 1 over 2 pi, f c, so i put values to get it 1 over 2 pi multiplied by if there is frequency and orchis 7.500 resistance, that is 25 nanooso. If a within it, we get x c of 8 fourty 8.8 kilo and of 4 x l, we have formula 2, pi, f l and we put values to get it so 1.5 is the frequency in the inductance has been given as 1 hund 50 microso? That x, equal to 0 point double 0707, ohms and z, is now equal to under root r square, that is 1 clomb square plus c, that is 0.0 double 0707, and that is minus 848.8 kilo. So that comes to be 848.8 kilo. Our power factor will be equal to r and we have r is 1 k or z that was found to be 848.8 kilo hm, so this results in power factor of 0.00 double 178. This is the power vector of the r l c series circuit, and this was our part. Now we proceed to part b in part b. We have to find the first difference. That is why, between voltage and the current, as you know, that power factor is equal to cos of i, and a power factor is already known that was 0.00 double 178 is equal to cosine of i y will be equal to miners of cosine into 0 point. Double 0 double 178 and this results in a 89.93 degree. So this is our first difference between the water current in the citied. In part c, we have to find a raise power that is p on frequency, latasunitsseven .5, so raise power is equal power factor multiplied by maximum power of the circuit that is in the power factor, was on point double 0 double 178. This or a raise power on this frequency. This is answer to our part c. In part, sorry, we have to find a raised power on resinous frequency, given that f is equal to. There is a resonance frequency in this condition. Our impudence will be equal to r and the power vector will be equal to 1 and then so power average power will be equal to 1, multiply maximum power, so it is equal to maximum power, togonatee and vector cancel out each other's affect. That is why maximum power is transferred to the circuit, and this frequency is known as resonance frequency nk you for watching.

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