A DC motor is supplied with 220 V and draws an armature current of 10 A.The armature resistance is 1 Ω. Then1.The back EMF developed in the motor is approximately 210 V.2.The electrical power converted into mechanical form is about 2100 W.3.When load increases, the armature current increases and speed decreases.4.The back EMF increases with increase in load current.Which of the above statements is/are correct?

Correct Answer: B - 1, 2 and 3 only

Quick Concept Explanation

In a DC motor, back EMF is generated by armature rotation and opposes the applied supply. Eb=V−IaRa=220−10×1=210 V. Converted electromagnetic power≈EbIa=210×10=2100 W. Key relation: E_b=V−I_aR_a; P_converted≈E_b I_a; T∝ΦI_a.

Exam focus: Higher load → initially lower speed/Eb → higher armature current. Common trap: Do not say Eb rises simply because load current rises at constant supply.

Statement-wise Verification

• Statement 1 - Correct.
  The back EMF developed in the motor is approximately 210 V.
  In a DC motor, back EMF is generated by armature rotation and opposes the applied supply.
• Statement 2 - Correct.
  The electrical power converted into mechanical form is about 2100 W.
  Converted electromagnetic power≈EbIa=210×10=2100 W.
• Statement 3 - Correct.
  When load increases, the armature current increases and speed decreases.
  Higher load → initially lower speed/Eb → higher armature current.
• Statement 4 - Incorrect.
  The back EMF increases with increase in load current.
  At constant terminal voltage, Eb=V−IaRa, so increased armature current tends to reduce back EMF unless other quantities change

Core Concept

In a DC motor, back EMF is generated by armature rotation and opposes the applied supply. The armature equation is V=Eb+IaRa (neglecting brush drop). Electromagnetic converted power is approximately EbIa. When mechanical load rises, speed and Eb initially fall, allowing Ia to rise and produce more torque.

Formula / Key Relationship

E_b=V−I_aR_a; P_converted≈E_b I_a; T∝ΦI_a.

Step-by-Step Check

Eb=V−IaRa=220−10×1=210 V. Converted electromagnetic power≈EbIa=210×10=2100 W.

Why the Other Options Are Wrong

Option A is incorrect because it includes incorrect statement(s) 4. Option C is incorrect because it omits correct statement(s) 1. Option D is incorrect because it includes incorrect statement(s) 4 and omits correct statement(s) 2, 3.

Exam Shortcut / Approach

Higher load → initially lower speed/Eb → higher armature current.

Common Mistake

Do not say Eb rises simply because load current rises at constant supply.

Quick Revision

DC Motor Back EMF and Converted Power is linked with Back EMF, DC Motors and Electrical Machines. Higher load → initially lower speed/Eb → higher armature current.