Neglecting all losses, the developed torque (T) of a DC separately excited motor, operating under co

Question

Neglecting all losses, the developed torque (T) of a DC separately excited motor, operating under constant terminal voltage, is related to its output power (P) as under:

Examoogle · Accepted Answer

$T \propto P$

Explanation: In a DC motor, the mechanical output power (P) is the product of the developed electromagnetic torque (T) and the angular speed (\omega). Under the condition of constant terminal voltage and neglecting losses, the motor maintains a speed-torque relationship such that torque is directly proportional to output power at a given operating point.

Working Principle: The mechanical power is expressed as $P = T 	imes \omega$. In a separately excited DC motor, $T = K_t \Phi I_a$ and $E = K_b \Phi \omega$. Given the terminal voltage $V$ is constant, $E = V - I_a R_a$. As power is $P = E I_a$, substituting $E$ shows that for a fixed field flux $\Phi$ and constant supply, the variation in torque tracks the mechanical power demand linearly when speed variations are considered in the steady-state equation.

Key Formulas: $P = T \omega$ — Relation between mechanical power, torque, and angular velocity; $T = K_t \Phi I_a$ — Expression for electromagnetic torque in a DC motor

Key Points: Mechanical power is defined as the rate of doing work, $P = T \omega$.; In a separately excited DC motor, torque is linearly proportional to the armature current.; Neglecting losses implies that the electrical input power is approximately equal to the developed mechanical power.; For a fixed speed or when considering the fundamental relationship, $T$ varies directly with $P$.

Advantages: Simple linear control characteristic; High starting torque capability

Disadvantages: Requires separate DC supply for field excitation; Brush and commutator maintenance

Applications: Industrial drives requiring variable speed; Conveyor systems and rolling mills

Additional Info: The assumption of 'neglecting losses' implies that the internal developed power equals the terminal electrical power $V I_a$.. Option A ($T \propto \sqrt{P}$) is incorrect as it does not follow the standard $P = T \omega$ linear relationship for rotating machinery.

Answer: B is correct — The mechanical power developed by a motor is the product of torque and angular speed; thus, at a constant speed, torque is directly proportional to power.

Exam Tip: Remember that $P = T\omega$ is a universal relationship for rotating machines; always look for the speed ($\omega$) constraint in DC motor problems.

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