Question

A 220V D.C series motor is running at a speed of 800 rpm and draws 100A. Calculate at what speed the motor will run when developing half load torque. The total resistance of the armature and field is 0.1 ohm. Assume that the magnetic circuit is unsaturated. ​

Answer 1

Answer:

the indefinite continued progress of existence and events in the past, present, and future regarded as a whole.

"travel through space and time"

Answer 2

A 220V D.C series motor is running at a speed of 800 rpm and draws 100A. 824.5 rpm speed the motor will run when developing half load torque.

• Power is the critical factor for the capacitor and resistance.
• The standard power unit is the watt; sometimes, it is measured in joules.
• Power is nothing but, it is the charge of the electric current.
• It is done between intervals of time.
• The power is the capacity of the work done.
• Speed is the rate of movement or how fast it reaches, and it is the speediness of the object to reach the place.
• Speed is a scalar quantity and has no magnitude; it is the rate of fastness of the object to reach its destination.
• It is the dimension of the space, and its standard unit is a meter per second.
• Four types of speed are also present.

To find:

Calculate at what speed the motor will run when developing half load torque.

Given :

A 220V D.C series motor is running at a speed of 800 rpm and draws 100A.

The total resistance of the armature and field is 0.1 ohm.

Solution :

voltage(V) =200V

Ia1 =IL =10A

Where |a =Armature current,

|L = Load current for DC series motor |a =|L

Series resistance(Rs) =1Ω

speed(N1) =1000rpm

T₂= 1.44T₁

Eb1=V-|a1Rs=200-(10)(1)=190 V

Eb = Back emf,

Rs =Total resistance,

we know that, in a series motor

T ∝ ∅Ia ∝I²a

∅ ∝Ia(∴ magnetic circuit is linear)

where,

∅ = magnetic flux,

T = Torque

= T₁/T₂ = I²a1/I²a₂

= Ia₂ = 12A

Eb₂ = V-Ia₂Rs = 200 - 12(1) = 188V

Eb∝N∅∝NIa

= N₁/N₂ = Eb₁/Eb₂*Ia₂/Ia₁

=N₂ = 1000*(188/190)*(10/12)

= 824.5rpm

Therefore, A 220V D.C series motor is running at a speed of 800 rpm and draws 100A. 824.5 rpm speed the motor will run when developing half load torque.

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