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A series DC motor draws 75 A when connected to a 600 V supply and rotates with a speed of 500 rpm. The motor has equal armature and series winding resistance of 0.5 N. Suppose the load torque is reduced to half (i.e., Tnew = 0.5Told), what would be the value of:

(a) The armature current. \
(b) The speed of rotation.

User MaGz
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1 Answer

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Final answer:

The new value of the armature current would be 37.5 A. The new speed of rotation cannot be calculated without knowing the value of K.

Step-by-step explanation:

The armature current of a series DC motor can be calculated using the equation:

Ia = (Eb - V) / Ra

Where Ia is the armature current, Eb is the back emf, V is the supply voltage, and Ra is the armature resistance.

In this case, we need to find the new value of the armature current when the load torque is reduced to half. Since the load torque is directly proportional to the armature current, if the load torque is halved, the armature current would also be halved.

Therefore, the new value of the armature current would be 0.5 * 75 A = 37.5 A.

In a series DC motor, the speed of rotation can be calculated using the equation:

N = (V - Ia * Ra) / K

Where N is the speed of rotation, V is the supply voltage, Ia is the armature current, Ra is the armature resistance, and K is a constant.

Since we know all the values except for the new speed of rotation, we can rearrange the equation to solve for N:

N = (V - Ia * Ra) / K

Plugging in the values, we get:

N = (600 V - 37.5 A * 0.5 N) / K

N = (600 V - 18.75 V) / K

N = 581.25 V / K.

Unfortunately, we do not have enough information to calculate the new speed of rotation without knowing the value of K.

User Grungetta
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