# A dc series motor driving and electric train faces ___

A dc series motor driving and electric train faces a constant power load.

It is running at rated speed and rated voltage.

If the speed has to be brought down to 0.25 p.u. the supply voltage has to be approximately brought down to

A. 0.75 p.u
B. 0.5 p.u
C. 0.25 p.u
D. 0.125 p.u

Share your understanding of this question with the correct explanation.

If a DC series motor driving an electric train with a constant power load is running at the rated speed and rated voltage, and the speed has to be brought down to 0.25 p.u. (per unit), the supply voltage has to be approximately brought down to 0.5 p.u. (per unit). Therefore, the correct answer is B: “0.5 p.u.”

Here’s a further explanation:

In a DC series motor, the speed of the motor is directly proportional to the supply voltage. In this scenario, the motor is initially running at the rated speed and rated voltage.

To bring down the speed of the motor to 0.25 p.u., which is 25% of the rated speed, the supply voltage needs to be reduced. The relationship between speed and voltage in a DC series motor is non-linear; it follows a power law relationship.

The speed (N) of a DC series motor is given by:

N ∝ V^α

where V represents the supply voltage and α is a constant exponent (usually between 0.8 and 1.2) determined by the motor design and characteristics.

Since we have a constant power load, reducing the speed to 0.25 p.u. means reducing the supply voltage by a certain factor. To determine the approximate value, we need to consider the power relationship between voltage and speed.

Power (P) in a DC motor is given by:

P = K * N * V

where K is a constant.

Since we have a constant power load, the power remains the same throughout. So, we can express the power relationship as:

P = K * N_initial * V_initial = K * N_final * V_final

Since we are reducing the speed to 0.25 p.u., we can write:

K * N_initial * V_initial = K * 0.25 * V_final

N_initial * V_initial = 0.25 * V_final

To find the approximate supply voltage required for the reduced speed, we can assume that the initial voltage is at the rated value (1 p.u.). Therefore:

1 * 1 = 0.25 * V_final

V_final ≈ 4 p.u.

However, the question asks for the approximate supply voltage brought down to p.u., which means we need to express the result in terms of p.u. Since 1 p.u. corresponds to the rated voltage, the supply voltage needs to be reduced to 0.5 p.u. to achieve the desired speed of 0.25 p.u.

Therefore, the correct answer is B: “0.5 p.u.”