Circuits

 

• Parallel circuits are circuits with components connected in parallel

• Series circuits are circuits with components connected in series

• AC circuits are circuits that have alternating current

• DC circuits are circuits that have direct current

• Active circuits are circuits that generate energy from electricity passing through them

• Passive circuits are circuits that don’t generate energy from electricity passing through them

State the difference between generator and alternator

Generator and alternator are two devices, which converts mechanical energy into electrical energy. Both have the same principle of electromagnetic induction, the only difference is that their construction. Generator persists stationary magnetic field and rotating conductor which rolls on the armature with slip rings and brushes riding against each other, hence it converts the induced emf into dc current for external load whereas an alternator has a stationary armature and rotating magnetic field for high voltages but for low voltage output rotating armature and stationary magnetic field is used.

What happens if DC supply is given to a transformer?

When dc supply is applied across the primary winding then a constant current of higher magnitude will flow through it due to low resistance of winding. It will setup a constant flux in the core

What is bypass capacitance?

                        Bypass capacitance is the property of the capacitor where it blocks DC and allows AC to pass through it. In the sense it allows DC to pass through the circuit.

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How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor Improvement  (Easiest way ever)

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