Electrical Engineering - RL Circuits - Discussion
Discussion Forum : RL Circuits - General Questions (Q.No. 16)
16.
Which of the following power factors results in less energy loss in an RL circuit?
Discussion:
17 comments Page 1 of 2.
Gokul Bhardwaj said:
7 years ago
Answer D is correct.
If the power factor is very less it means there is a large value of inductance in the circuit. It causes an increase in reactance and a decrease in current. If current decreases then power loss is also decreased for constant resistance. However, the disadvantage of keeping the power factor low is that active power is decreased. Thus to do useful work active power must be increased and power factor must be enhanced.
If the power factor is very less it means there is a large value of inductance in the circuit. It causes an increase in reactance and a decrease in current. If current decreases then power loss is also decreased for constant resistance. However, the disadvantage of keeping the power factor low is that active power is decreased. Thus to do useful work active power must be increased and power factor must be enhanced.
Ravi Koshiya said:
9 years ago
Energy losses = I^2 * R, clearly, it's seen then it depends on are.
But, as we talk about R-L circuit, inductor comes into the picture makes energy losses depends on z instead of are only. So, if we increase XL than there will be a decrease in P.F. as well as in current. It's clear that in RL circuit low P.F. Comes with fewer energy losses.
But, as we talk about R-L circuit, inductor comes into the picture makes energy losses depends on z instead of are only. So, if we increase XL than there will be a decrease in P.F. as well as in current. It's clear that in RL circuit low P.F. Comes with fewer energy losses.
NEERAJ KUMAR said:
5 years ago
Answer D is correct.
Because we know that energy loss means here works reactive Power [VI SIN (θ) ], and in sine function, if we increase the value of θ then the value will be increased, which means here we have to choose the minimum value of Power factor.
Because we know that energy loss means here works reactive Power [VI SIN (θ) ], and in sine function, if we increase the value of θ then the value will be increased, which means here we have to choose the minimum value of Power factor.
Ajit kumar yadav said:
5 years ago
Power factor = true power/ apparent power.
Therefore power factor inversely proportional to apparent power.
Therefore to minimise apparent power which consists of reactive power that causes more loss, we have to decrease power factor.
Therefore power factor inversely proportional to apparent power.
Therefore to minimise apparent power which consists of reactive power that causes more loss, we have to decrease power factor.
Kartik Pawar said:
10 years ago
I think it should be option B because at unity pf current is less so losses will be less. But since unity pf is not possible in RL circuit so pf should be close to unity. So option B.
Vijay said:
9 years ago
Answer d is correct, because at pure resistance power loss is high and power factor would be 1. To less power loss in RL circuit, Power factor should be as low as possible.
Deependra Kr said:
7 years ago
loss in RL crt= v^2/(R^2+wL^2)*R.
Loss=v^2/(R^2+wL^2)^1/2*cosθ.
=> loss proportional to pf.
So, the less value of pf is recommended.
Loss=v^2/(R^2+wL^2)^1/2*cosθ.
=> loss proportional to pf.
So, the less value of pf is recommended.
Awais said:
7 years ago
P = VI*pf.
Power is directly proportional to power factor. Therefore for minimum power loss pf should be minimum.
Power is directly proportional to power factor. Therefore for minimum power loss pf should be minimum.
Ahsan said:
6 years ago
D is correct.
PF= R/Z.
Which means greater PF has greater has greater R consequently greater losses.
PF= R/Z.
Which means greater PF has greater has greater R consequently greater losses.
Madan C said:
9 years ago
At zero pf current will be more, so that copper losses increases, the correct answer is unity pf.
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