Electrical Engineering - RL Circuits - Discussion

Discussion :: 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

[A]. 564
[B]. 0.564
[C]. 6.76
[D]. 55.7

Answer: Option B

Explanation:

No answer description available for this question.

Sanjeeb said: (Aug 31, 2011)  
COSQ=R/Z=1.5/1.5+2.2=0.564

Manfrizy said: (Mar 11, 2012)  
R/Z = 0.405 not 0.564 as you put it Sanjeeb

Karimulla said: (Mar 19, 2012)  
Generally power factor i.e. cos(angle b/w v,i) always less than one.

Yousuf said: (Jun 1, 2012)  
|z|=sqrt(1500^2+2200^2)=2663
Angle=inverse tan(2200\1500)=55.7
Power Factor=Cos(55.7)=0.564

Ryan said: (Feb 18, 2013)  
Z = sqrt(1500^2+2200^2) = 2663.

pf = R/Z = 1500/2663 = 0.564.

Amit Kumar Jaiswal said: (Nov 2, 2013)  
POWER FACTOR = (R/Z).

Z = SQRT(1.5^2 +2.2^2).

Z = 2.662.

P.F. = 1.5/2.662 = .563.

Shaila Kedar said: (Feb 21, 2014)  
Power factor is always less than 1.

Noel Zamora said: (Jun 5, 2014)  
tan 2.2/1.5 = 1.4667, arc tan 1.4667 = 55.71 deg.,

Therefore cos 55.71 deg. = 0.564 or pf = 0.564(lagging).

Hadiro Tafoua said: (Mar 9, 2015)  
Using all method above we get close intervals of answers. Since it all corresponds to electric quantities.

Shenoy said: (Jun 27, 2015)  
Power factor always less than 1 use common sense, so simple.

Venkatesh said: (Aug 28, 2015)  
R = 1.5.
Xl = 2.2.

Z = Root of R square+xl square.
Cos = R/Z.

Pavan said: (Oct 21, 2015)  
Power factor indicates efficiency of the circuit.

G1Kumar said: (Jul 22, 2016)  
z = (r^2 + x^2)^(1/2).
Ccos = r/z.

Vicky said: (Oct 4, 2016)  
cos(q) can never be greater than 1.

Saipriya said: (Nov 25, 2016)  
Describe or view the process of the answer.

Manoj Khodade said: (Sep 5, 2017)  
How to solve? please details.

Arpana Singh said: (Jun 15, 2018)  
R/Z=0.405 not 0.564.

Siddharth said: (Mar 7, 2019)  
Power factor = R ÷ Z.

So, R = 1.3k ohm = 1.3 * 10^3 =1300 ohm,
inductive reactians FL = 2.2k ohm =2.2 * 10^3,
= 2200 ohm.
&. Z = (?).

Z = √(R^2 + FL^2),
Z = √(1500^2 + 2200^2),
Z= 2663.

& Power factor = R ÷ Z,
= 1500 ÷ 2663,
= 0.5632.
So, ans is (0.564).

Kalyan G said: (May 3, 2020)  
Z=(r^2+XL^2)^1/2=2663.
P.f = cosq = r/z
0.564.

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