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


Discussion:
20 comments Page 1 of 2.
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.
The main simple thing we need to see here is the power factor should not be greater than 1.
(3)
Kalyan G said:
5 years ago
Z=(r^2+XL^2)^1/2=2663.
P.f = cosq = r/z
0.564.
P.f = cosq = r/z
0.564.
(1)
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).
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)
Arpana singh said:
7 years ago
R/Z=0.405 not 0.564.
(1)
Manoj Khodade said:
8 years ago
How to solve? please details.
Saipriya said:
9 years ago
Describe or view the process of the answer.
Vicky said:
9 years ago
cos(q) can never be greater than 1.
(1)
G1KUMAR said:
9 years ago
z = (r^2 + x^2)^(1/2).
Ccos = r/z.
Ccos = r/z.
Pavan said:
10 years ago
Power factor indicates efficiency of the circuit.
VENKATESH said:
1 decade ago
R = 1.5.
Xl = 2.2.
Z = Root of R square+xl square.
Cos = R/Z.
Xl = 2.2.
Z = Root of R square+xl square.
Cos = R/Z.
(1)
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