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
564
0.564
6.76
55.7
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
20 comments Page 1 of 2.

Siddharth said:   7 years ago
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).
(3)

Noel Zamora said:   1 decade ago
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:   1 decade ago
Using all method above we get close intervals of answers. Since it all corresponds to electric quantities.

Pavan said:   2 years ago
@All.

The main simple thing we need to see here is the power factor should not be greater than 1.
(3)

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

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.

Karimulla said:   1 decade ago
Generally power factor i.e. cos(angle b/w v,i) always less than one.

VENKATESH said:   1 decade ago
R = 1.5.
Xl = 2.2.

Z = Root of R square+xl square.
Cos = R/Z.
(1)

Ryan said:   1 decade ago
Z = sqrt(1500^2+2200^2) = 2663.

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

Shenoy said:   1 decade ago
Power factor always less than 1 use common sense, so simple.
(1)


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