# How to calculate current motor value in single phase motor?

:1: There is an electric motor that used to drive conveyor machine. The electric motor has specifications: voltage is 200 Volt single phase, frequency is 50 Hz, cos phi is 0.82 and real power on the name plate is 330 Watt. How many

:catat: Answer:

V = 200 Volt

f = 50 Hz

Cos phi = 0.82

P = 330 W

I = ...?

===================================

The formula of real power electricity in single phase motor

P = V I Cos phi . . . . . (Watt)

:pin: then

====================================

I = 330 / (200 * 0.82)

I = 2 A.

:2: In the chocolate factory has an electric motor that has specifications: voltage is 220 Volt single phase, frequency is 50 Hz, cos phi is 0.78 and real power on the name plate is 2.2 kW. How many

:catat: Answer:

V = 220 Volt

f = 50 Hz

Cos φ = 0.78

P = 2.2 kW →

I = ...?

I = P / (V Cos φ)

I = 2200 / (220 * 0.78)

I = 12.82 A.

:3: The electric motor has specifications: voltage is 400 Volt single phase, frequency is 60 Hz, Cos φ is 0.85 Ampere and real power on the name plate is 6 HP. How many

:catat: Answer:

V = 400 Volt

f = 60 Hz

Cos φ = 0.85

P = 6 HP →

I = ...?

I = P / (V Cos φ)

I = 4476 / (400 * 0.85)

I = 13.165 A.

**current of electricity**will flow into the motor?:catat: Answer:

V = 200 Volt

f = 50 Hz

Cos phi = 0.82

P = 330 W

I = ...?

===================================

The formula of real power electricity in single phase motor

P = V I Cos phi . . . . . (Watt)

:pin: then

**The formula of current electricity in single phase motor**

I = P / (V Cos phi) . . . . . (Ampere)

I = P / (V Cos phi) . . . . . (Ampere)

====================================

I = 330 / (200 * 0.82)

I = 2 A.

:2: In the chocolate factory has an electric motor that has specifications: voltage is 220 Volt single phase, frequency is 50 Hz, cos phi is 0.78 and real power on the name plate is 2.2 kW. How many

**current of electricity**will flow into the motor?

:catat: Answer:

V = 220 Volt

f = 50 Hz

Cos φ = 0.78

P = 2.2 kW →

**convert kW into W**→ 2.2 * 10

^{3}W = 2200 W

I = ...?

I = P / (V Cos φ)

I = 2200 / (220 * 0.78)

I = 12.82 A.

:3: The electric motor has specifications: voltage is 400 Volt single phase, frequency is 60 Hz, Cos φ is 0.85 Ampere and real power on the name plate is 6 HP. How many

**current of electricity**will flow into the motor?

:catat: Answer:

V = 400 Volt

f = 60 Hz

Cos φ = 0.85

P = 6 HP →

**convert HP into W**→ 6 * 746 W = 4476 W

I = ...?

I = P / (V Cos φ)

I = 4476 / (400 * 0.85)

I = 13.165 A.

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