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 Given:

A DC shunt motor with armature resistance of 0.12 Ohm is fed with a 220 V supply. It is started using a variable resistance starter and starting characteristics are provided with a minimum current of 162 A and a maximum current of 270 A.

Step 2

We know that at the instant of starting, the back emf induced in the armature is zero. At this instant, if no starter is used, there will be a maximum current in the armature. This maximum current flowing in the armature circuit is given by

Electrical Engineering homework question answer, step 2, image 1

Answer (a): The maximum current that can flow through the armature if there is no starting resistance is present is 1833.33 A .

Step 3

When a starter is connected in series of armature circuit, the starting current will be limited. The armature current will be reduced from 1833.33 A to 270 A and as its speed increases, back emf increases. So, the armature current will also be reduced. Here, the minimum armature current flowing in the circuit with the starter is 162 A.

Answer (b): 162 A .

Step 4

From the starting characteristics, the ratio of upper limit and lower limit of current can be written as

Electrical Engineering homework question answer, step 4, image 1

The total armature circuit resistance (including starter resistance) at the instant of starting is

Electrical Engineering homework question answer, step 4, image 2

The number of starting resistor section (n) can be calculated as

Electrical Engineering homework question answer, step 4, image 3

For an integral choice, n = 5.

Answer (c): .

Step 5

At the instant of starting, the total armature resistance (including starting resistance) calculated is 0.8148 Ohm. The required starter resistance is

Electrical Engineering homework question answer, step 5, image 1

Answer (d): 0.69 Ohm .

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