# 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.

Pavan said:   5 months ago
@All.

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

Kalyan G said:   4 years ago
Z=(r^2+XL^2)^1/2=2663.
P.f = cosq = r/z
0.564.

Siddharth said:   5 years ago
Power factor = R Ã· 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 Ã· Z,
= 1500 Ã· 2663,
= 0.5632.
So, ans is (0.564).
(3)

Arpana singh said:   6 years ago
R/Z=0.405 not 0.564.
(1)

Manoj Khodade said:   7 years ago
How to solve? please details.

Saipriya said:   8 years ago
Describe or view the process of the answer.

Vicky said:   8 years ago
cos(q) can never be greater than 1.

G1KUMAR said:   8 years ago
z = (r^2 + x^2)^(1/2).
Ccos = r/z.

Pavan said:   9 years ago
Power factor indicates efficiency of the circuit.

VENKATESH said:   9 years ago
R = 1.5.
Xl = 2.2.

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

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