A 120-V, 60-Hz source supplies current to a 1-µF capacitor, a 7.036-H inductor, and a 1-O...1 answer below »

A 120-V, 60-Hz source supplies current to a 1-µF capacitor, a 7.036-H inductor, and a 1-O resistor, all wired in parallel as shown in the figure below.

a. What are the reactances (O) of the capacitor and the inductor? b. What is the rms current (A) through each of the three load components? c. Express the impedance Z of each load component in polar vector form with a magnitude Z(O) and a phase angle f.

d. Write the currents I through each of the three load components in the form of phasors, with a magnitude Irms(A) and a phase angle f. See equation (3.48b) of Masters 2nd edition for a useful relation regarding vector division.

e. What is the total rms current (A) delivered by the source? To answer this, you may find it convenient to convert the phasors for the currents in part (d) to complex notation; i.e., I = a + jb. f. What is the power factor? Why does it have this value for this particular circuit?

Here all the solution are attached Reactance of Capacitor(Xc) =1/ ?C = 1/2pfC Where C = Capacitance, ? = angular frequency, f= frequency
Here given f = 60 Hz, V=120 V, C = 1uF, L = 7.036 H, R =1?
So Xc =1/(2*p*60*1*10-6) = 2.653 K ? Reactance of Inductor(XL) =?L = 2pfL Where L = Inductance, ? = angular frequency, f= frequency
Here given f = 60 Hz, V=120 V, C = 1uF, L = 7.036 H, R =1?
So XL =(2*p*60*7.036) = 2.653 K ? I_r=Vrms/R=(120v2)/1=169.68 Amp
I_l=Vrms/X_l =(120v2)/2.653K=63.96 mA
I_C=Vrms/X_C =(120v2)/2.653K=63.96 mA 1/z_eq =1/z_r +1/z_l +1/z_c =1/R+1/j?L+j?C=1/R+j(?C-1/?L)=1+j(2.653-2.653) 1/z_eq =1
z_eq=1+j0
tanØ=-R(X_c-1/X_l )=0
Ø=?tan?^(-1) (-R(X_c-1/X_l ))=?tan?^(-1) 0
Phase angle(Ø)=0 I_r=Vrms/R=(120v2)/1=169.68 Amp
I_l=Vrms/X_l =(120v2)/2.653K=63.96 mA
I_C=Vrms/X_C =(120v2)/2.653K=63.96 mA Rms current I_rms=V_rms [1/R^2 +(?C- 1/?L )^2 ]^(1/2) Irms = Vrms*1 = 120 * 1.414 = 169.68Amp Because Xc = XL So Irms = 1/R Power factor is defined as the ratio of True Power to Apparent Power
Power Factor=TruePower/(Apparent Power) Power Factor=TruePower/(Apparent Power)=(I^2 R)/(I^2 Z)=R/Z=R/R=1

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Here all the solution are attached Reactance of Capacitor(Xc) =1/ ?C = 1/2pfC Where C = Capacitance, ? = angular frequency, f= frequency
Here given f = 60 Hz, V=120 V, C = 1uF, L = 7.036 H, R =1?
So Xc =1/(2*p*60*1*10-6) = 2.653 K ? Reactance of Inductor(XL) =?L = 2pfL Where L = Inductance, ? = angular frequency, f= frequency
Here given f = 60 Hz, V=120 V, C = 1uF, L = 7.036 H, R =1?
So XL =(2*p*60*7.036) = 2.653 K ? I_r=Vrms/R=(120v2)/1=169.68 Amp
I_l=Vrms/X_l =(120v2)/2.653K=63.96 mA
I_C=Vrms/X_C =(120v2)/2.653K=63.96 mA 1/z_eq =1/z_r +1/z_l +1/z_c =1/R+1/j?L+j?C=1/R+j(?C-1/?L)=1+j(2.653-2.653) 1/z_eq =1
z_eq=1+j0
tanØ=-R(X_c-1/X_l )=0
Ø=?tan?^(-1) (-R(X_c-1/X_l ))=?tan?^(-1) 0
Phase angle(Ø)=0 I_r=Vrms/R=(120v2)/1=169.68 Amp
I_l=Vrms/X_l =(120v2)/2.653K=63.96 mA
I_C=Vrms/X_C =(120v2)/2.653K=63.96 mA Rms current I_rms=V_rms [1/R^2 +(?C- 1/?L )^2 ]^(1/2) Irms = Vrms*1 = 120 * 1.414 = 169.68Amp Because Xc = XL So Irms = 1/R Power factor is defined as the ratio of True Power to Apparent Power
Power Factor=TruePower/(Apparent Power) Power Factor=TruePower/(Apparent Power)=(I^2 R)/(I^2 Z)=R/Z=R/R=1