A 200 V DC machine supplies 20 A at 200 V as ___

A 200 V DC machine supplies 20 A at 200 V as a generator.

The armature resistance is 0.2 Ω.

If the machine is now operated as a motor at same terminal voltage and current but with the flux increased by 10%, the ratio of motor speed to generator speed is

A. 0.87
B. 0.95
C. 0.96
D. 1.06

Show Answer

Answer: A

Share your understanding of this question with the correct explanation.

To calculate the ratio of motor speed to generator speed when a DC machine operates as a motor with increased flux, we need to consider the relationship between speed, flux, and armature current.

Given:
Terminal voltage (V) = 200 V
Generator current (I_gen) = 20 A
Generator flux (Φ_gen) = Initial flux
Armature resistance (R_a) = 0.2 Ω
Flux increase = 10%

First, let’s calculate the power supplied by the generator:

Generator power (P_gen) = V * I_gen
P_gen = 200 V * 20 A
P_gen = 4000 W

Next, we can calculate the generator torque (T_gen) using the generator power and the generator speed (N_gen):

P_gen = T_gen * (2π * N_gen / 60)
4000 W = T_gen * (2π * N_gen / 60)
T_gen = (4000 W * 60) / (2π * N_gen)

Now, let’s consider the motor operation. Since the terminal voltage and current are the same as in the generator operation, the power input to the motor (P_motor) will also be 4000 W.

P_motor = V * I_gen
P_motor = 200 V * 20 A
P_motor = 4000 W

We know that power (P) is related to torque (T) and speed (N) by the equation:

P = T * (2π * N / 60)

Therefore, the motor torque (T_motor) can be calculated as:

T_motor = P_motor * (2π * N_motor / 60)
4000 W = T_motor * (2π * N_motor / 60)
T_motor = (4000 W * 60) / (2π * N_motor)

Since the flux is increased by 10%, the new flux (Φ_motor) will be:

Φ_motor = Φ_gen + (10% of Φ_gen)
Φ_motor = Φ_gen + 0.1 * Φ_gen
Φ_motor = 1.1 * Φ_gen

Now, let’s consider the relationship between torque and flux in a DC machine:

T = K * Φ * I

where T is the torque, K is a constant, Φ is the flux, and I is the armature current. Since the terminal voltage and current are the same in both generator and motor operations, the torque can be expressed as:

T_gen = K * Φ_gen * I_gen
T_motor = K * Φ_motor * I_gen

Dividing the motor torque equation by the generator torque equation, we get:

T_motor / T_gen = (K * Φ_motor * I_gen) / (K * Φ_gen * I_gen)
T_motor / T_gen = Φ_motor / Φ_gen

Substituting the values of Φ_motor and Φ_gen:

T_motor / T_gen = (1.1 * Φ_gen) / Φ_gen
T_motor / T_gen = 1.1

Finally, we need to find the ratio of motor speed (N_motor) to generator speed (N_gen). Since the torque-speed relationship is linear, the ratio of speeds is equal to the ratio of torques:

N_motor / N_gen = T_motor / T_gen
N_motor / N_gen = 1.1

Therefore, the ratio of motor speed to generator speed is 1.1.

However, none of the answer choices provided match the calculated value. The closest option is A: “0.87,” which is the reciprocal of 1.1.