3) In an RL circvit,a 110-V Crr), 6o-H Source ir ia Series with a So.om H cnduct

Question

3) In an RL circvit,a 110-V Crr), 6o-H Source ir ia Series with a So.om H cnductor and a 3o.o-l res:stor. FiAd a) The rms current in the cireuit b) T he power facter and pawer contomed by the lead c) The size of capaci tor that you must add fo that the power factor is 1 mean ing that the currenf in the load iS in phase with the source Voltage to maxi mize effici ent energy tranrfer, d) with the new capacitor, how much can the input (supply Vo ltqqe be re duced while maintainina the previous Power con sumption

Explanation / Answer

3)a)RMS current in the circuit = RMS voltage/Resistance

=120/30

= 4A

b)Ztotal = (R^2 + (wL)^2)^0.5

=(30^2 + (2*pi*60*50*10^-3)^2)^0.5

=35.43 ohm

Current in the circuit = V/Z

=130/35.43

=3.669 A

so, power consumed by the load = I^2*R

=3.669^2*30

=403.847 W

Power factor = R/Z

=30/35.43

=0.8467

c)Zcl = 0

or wL = 1/wC

or C = 1/(w^2*L)

=1/((2*pi*60)^2*0.05)

=1.4*10^-4 F

d)let the reduced voltage be V.so,

V^2/30 = 403.847

or V=110.07 V

so it can be reduced by 9.3 V

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