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Electric Motor Control

10th Edition
Herman
Publisher: CENGAGE L
ISBN: 9781133702818

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BuyFindarrow_forward

Electric Motor Control

10th Edition
Herman
Publisher: CENGAGE L
ISBN: 9781133702818
Chapter 44, Problem 3SQ
Textbook Problem
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If an armature has a resistance of 0.5 ohm, what series resistance must be used to limit the current to 30 amperes across a 230-volt circuit at the instant of starting? What is the power rating?

To determine

Find the series resistance required to limit current to 30A across a 230V circuit at the instant of starting, and also find the power rating of the circuit.

Explanation of Solution

Given data:

The armature resistance (Ra) is 0.5Ω.

The armature current (I) is 30A.

The line voltage (E) is 230V.

Formula used:

Write the expression to find the series resistance of the circuit.

R=EEcIRa (1)

Here,

E is the line voltage in volts,

I is the armature current in amperes,

Ec is the counter emf in volts, and

Ra is the armature resistance in ohms.

Write the expression to find power rating of the circuit.

P=I2R (2)

Here,

I is the armature current in amperes,

R is the series resistance in ohms.

Calculation:

When a motor is at standstill the rotating armature is in the stationary position and therefore, no counter emf will be generated, that is, Ec=0.

Substitute 230V for E, 0 for Ec, 30A for I, and 0.5Ω for Ra in Equation (1) to find R.

R=(230V0)30A0.5Ω=7.66VA0.5Ω (3)

Substitute unit Ω for VA in equation (3) to find resistance R in ohms

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