Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 1)
1.
A 1.5 k
resistor and a coil with a 2.2 k inductive reactance are in series across an 18 V ac source. The power factor is
resistor and a coil with a 2.2 k inductive reactance are in series across an 18 V ac source. The power factor isDiscussion:
20 comments Page 2 of 2.
SANJEEB said:
1 decade ago
COSQ=R/Z=1.5/1.5+2.2=0.564
Pavan said:
10 years ago
Power factor indicates efficiency of the circuit.
G1KUMAR said:
9 years ago
z = (r^2 + x^2)^(1/2).
Ccos = r/z.
Ccos = r/z.
AMIT KUMAR JAISWAL said:
1 decade ago
POWER FACTOR = (R/Z).
Z = SQRT(1.5^2 +2.2^2).
Z = 2.662.
P.F. = 1.5/2.662 = .563.
Z = SQRT(1.5^2 +2.2^2).
Z = 2.662.
P.F. = 1.5/2.662 = .563.
Saipriya said:
9 years ago
Describe or view the process of the answer.
Manoj Khodade said:
8 years ago
How to solve? please details.
Ryan said:
1 decade ago
Z = sqrt(1500^2+2200^2) = 2663.
pf = R/Z = 1500/2663 = 0.564.
pf = R/Z = 1500/2663 = 0.564.
Yousuf said:
1 decade ago
|z|=sqrt(1500^2+2200^2)=2663
Angle=inverse tan(2200\1500)=55.7
Power Factor=Cos(55.7)=0.564
Angle=inverse tan(2200\1500)=55.7
Power Factor=Cos(55.7)=0.564
Karimulla said:
1 decade ago
Generally power factor i.e. cos(angle b/w v,i) always less than one.
Manfrizy said:
1 decade ago
R/Z = 0.405 not 0.564 as you put it Sanjeeb
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