# Electric Drives Questions and Answers – Dynamics – Load Torques that Depend on the Path or Position Taken by the Load During Motion

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This set of Electric Drives Assessment Questions and Answers focuses on “Dynamics – Load Torques that Depend on the Path or Position Taken by the Load During Motion”.

1. What is the empirical formula for the tractive force required to overcome curve resistance? (W-the weight of the body, R – radius of curvature)
a) 710×W÷R
b) 700×W÷R
c) 720×W÷R
d) 750×W÷R

Explanation: Fc= 700×W÷R is the tractive force required to overcome curve resistance where W is the weight of the body in Kg and R be the radius of curvature in meters.

2. Force resisting the upward motion of a body on an inclined plane is given by (alpha – the angle of inclination, W- the weight of the body).
a) F = W×sin(alpha)
b) F = W×cosec(alpha)
c) F = W×sec(alpha)
d) F = W×cos(alpha)

Explanation: When a body is moving upward on an inclined plane its weight can be resolved in two perpendicular components that are W×sin(alpha) and W×cos(alpha). W×cos(alpha) is the component that is opposite to normal of the inclined plane and W×sin(alpha) is the component that opposes the upward motion of the body.

3. The unit of the torque is ______
a) N-m
b) N-m2
c) N-m/sec
d) N-Hz

Explanation: Torque is defined as the vector product of displacement and force. The unit of the force is Newton(N) and of the displacement is a meter (m) so the unit of torque in N-m.
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4. Calculate the value of the torque when 10 N force is applied perpendicular to a 10 m length of rod fixed at the center.
a) 200 N-m
b) 300 N-m
c) 100 N-m
d) 400 N-m

Explanation: Torque can be calculated using the relation T = (length of rod) × (Force applied) = r×F×sin90. F is given as 10 N and r is 10 m then torque is 10×10 = 100 N-m. (the angle between F and r is 90 degrees)

5. What is the dimensional formula for torque?
a) [ML2T-2]
b) [MLT-2]
c) [M1L2T-3]
d) [LT-2]

Explanation: Torque is a vector product of force and displacement. Dimensional formula for force is [MLT-2] and displacement is [L] so dimensional formula for torque is [MLT-2] [L] = [ML2T-2].

6. Buck converter is used to _________
a) Step down the voltage
b) Step up the voltage
c) Equalize the voltage
d) Step up and step down the voltage

Explanation: The output voltage of the buck converter is Vo = D×Vin. The value of the duty cycle is less than one which makes the Vo < Vin. The buck converter is used to step down voltage. Vin is a fixed voltage and Vo is a variable voltage.

7. If the starting torque of the motor is less than the load torque, the motor will fail to start.
a) True
b) False

Explanation: J×d(w)÷d(t) = Motor torque – Load torque is the dynamic equation of motor. If starting torque (motor torque) is less than the load torque then d(w)÷d(t) <0, acceleration <0 so the motor will decelerate and fails to start.

8. Torque is a scalar quantity.
a) True
b) False

Explanation: Scalar quantity has only magnitude whereas vector quantity has both directions and magnitude. Torque is a force applied on a body perpendicularly. As the force is a vector quantity, the torque must be treated as a vector quantity.

9. 250V, 15A, 1100 rpm separately excited dc motor with armature resistance (Ra) equal to 2 ohms. Calculate back emf developed in the motor when it operates on half of the full load. (Assume rotational losses are neglected)
a) 210V
b) 240V
c) 230V
d) 235V

Explanation: Back emf developed in the motor can be calculated using the relation Eb = Vt-I×Ra. In question, it is asking for half load, but the data is given for full load so current becomes half of the full load current = 15÷2 = 7.5 A. 250V is terminal voltage it is fixed so Eb = 250-7.5×2 = 235V.