AC Resistance
In Simple words, Resistance in AC circuits is called Impedance. Or
The Overall resistance (Resistance, Inductive reactance and Capacitive reactance) in AC circuits is called Impedance (Z).
Explanation:
When AC Current pass through a wire (resistor, inductor), then current produces a magnetic field across that wire which opposes the flow of AC Current in it along with the resistance of that wire. This oppose cause is called Inductance or Inductance is the property of Coil (or wire) due to which opposes any increase or decrease of current or flux through it. Also, we know that inductance is only exist in AC because the magnitude of current continuously changing
Inductive Reactance XL, is the property of Coil or wire in an AC circuit which opposes the change in the current. The unit of Inductive reactance is same as Resistance, capacitive reactance i.e. Ohm (?) but the representative symbol of capacitive reactance is XL.
Likewise,
Capacitive Reactance in a capacitive circuit is the opposition to current flow in AC circuits only. The unit of capacitive reactance is same as Resistance, Inductive reactance i.e. Ohm (?) but the representative symbol of capacitive reactance is XC.
Measuring AC Resistance
Hi
Example: 1
A 3 Phase, 5 kW Induction Motor has a P.F (Power factor) of 0.75 lagging. What size of Capacitor in kVAR is required to improve the P.F (Power Factor) to 0.90?
Solution #1 (By Simple Table Method)
Motor Input = 5kW
From Table, Multiplier to improve PF from 0.75 to 0.90 is .398
Required Capacitor kVAR to improve P.F from 0.75 to 0.90
Required Capacitor kVAR = kW x Table 1 Multiplier of 0.75 and 0.90
= 5kW x .398
= 1.99 kVAR
And Rating of Capacitors connected in each Phase
1.99/3 = 0.663 kVAR
Solution # 2 (Classical Calculation Method)
Motor input = P = 5 kW
Original P.F = Cosθ1 = 0.75
Final P.F = Cosθ2 = 0.90
θ1 = Cos-1 = (0.75) = 41°.41; Tan θ1 = Tan (41°.41) = 0.8819
θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843
Required Capacitor kVAR to improve P.F from 0.75 to 0.90
Required Capacitor kVAR = P (Tan θ1 – Tan θ2)
= 5kW (0.8819 – 0.4843)
= 1.99 kVAR
And Rating of Capacitors connected in each Phase
1.99/3 = 0.663 kVAR
- Badges
- Science topic
- Similar topics
- Applied Mathematics
- Calculations