What is the impedance of a 2H inductor in series with a 1K r

What is the impedance of a 2H inductor in series with a 1K resistor at a frequency of 200 Hz? What must the frequency be for a 10uf capacitor to have a reactance of 1000 ohms? An inductor has a reactance of Ohms. What is the total impedance in the polar form if it is connected in series with a 10 Ohm resistor? What will be the total impedance in polar form if the resistor and inductor in part 3 are connected in parallel? What is the total impedance in polar form of a 30 Ohm resistor, an inductor with 60 Ohms reactance and a capacitor with 20 Ohms reactance are connected in series? What will be the current if a 20 Ohm resistor and an inductor with a reactance of 10 Ohms are connected in series to a 120 volt (AC) source? Find Z in the polar form for Z = 12 +j8. Find Z in the rectangular form if the absolute value of Z is 20 and the angle is 60 degrees. Find Z in the polar form for Z = (8 + j6)/(10 + j12). Find Z in the polar form for Z = (6 + j6)(4 + j3). What is meant by the RMS current. How is it found? Give the value of the real and imaginary power for a 120 volt source at a phase angle of zero connected to an impedance of Z = 40 + j20 Ohms. What is the power factor? A.1 H inductor and 1.1 micro-farad capacitor are connected in series. What will be the resonance frequency? What will be the current if a 10 ohm resistor is connected in series along with the elements in problem 13 if the voltage is 120sin(377t)?

Solution

1) Z=R+jXL=(1000+j2513.2) ohms

2) XC=1000 ohms, C=10*10^-6 F

1/(WC)=XC

XC=1/(2*3.14*f*C)

f=1/(1000*10*10^-6*2*3.14)=15.915 hz

3) Z=(10+j10) ohms

Z=14.142<45 ohms

4) Z=[(10*j10)/(10+j10)]=7.07<45 ohms

5) Z=30+j(60-20)=(30+j40) ohms

6) current=120<0/(20+j10)=5.366<-26.56 amps

7) z=(12+j8)=14.42<33.69 ohms

8) z=20<60=(10+j17.32) ohms

9) z=(8+j6)(10+j12)=156.204<87.06 ohms

10) z=(6+j6)(4)+j3)=42.426<81.869 ohms

12) p=v^2/z=(120<0)^2/(40+j30)=288< -36.86 watts

power factor=cos(36.86)=0.8

13) fc=1/(2*3.14*square root of(L*C))=1/(2*3.14*square root of(0.1*0.1*10^-6))=1591.54 hz

14) i=[120<0/(10+j(377*0.1-1/(377*0.1*10^-6))]=4.53*10^-3<89.9 amps.