# How does at low power factor impedance (p.f=R/Z) (if R is constant ) increases while current also increases?

Impedance is equivalent resistance of circuit having resistive and reactive components so if we keep R constant and increase inductance than Z will increase and power factor will going to decrease as power factor can also be stated as ratio of R and Z. (R/Z) So here comes a contradiction that increasing impedance increasing the current.

• Who says it does? Jun 27 '19 at 7:06
• What do you mean by this. I am asking at low power factor current increases but as per formula p.f=V/Z the increase in impedance increases power factor. Jun 27 '19 at 8:41
• Why do you say that increasing Z, increases current? Sure, R/Z decreases, but R/Z is not an impedance so doesn't feature in Ohm's law
– Chu
Jun 27 '19 at 9:59
• R/Z decreases but if you keep resistive load constant then Z increases which follows decrease in R/Z. Suppose if i connect 100 W bulb with the wire having some inductance so the impedance would be Z=(R^2+Xj^2)^0.5 so if you keep R constant and increases X then Z increases which lowers the power factor. Jun 28 '19 at 4:14
• You need to provide more information because as current increases, active power increases, but so does apparent power so we are back to ratios. What type of circuit are you referencing? Series RL, parallel RL, motor? Jul 2 '19 at 4:11

$$I = \frac {V_S} {Z}$$