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.
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)
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.
Vicky said:
9 years ago
cos(q) can never be greater than 1.
(1)
Saipriya said:
9 years ago
Describe or view the process of the answer.
Manoj Khodade said:
8 years ago
How to solve? please details.
Arpana singh said:
7 years ago
R/Z=0.405 not 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)
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)
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)
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