Question

100 V, 2 A, 90 rpm separately excited dc motor with armature resistance (Ra) equal to 8 ohms.  Calculate back emf developed in the motor when it operates on 3^th/4 of the full load. (Assume rotational losses are neglected)

(a) 100 V

(b) 87 V

(c) 88 V

(d) 90 V

The question was asked by my school principal while I was bunking the class.

My question comes from Methods of Starting Electric Motors topic in chapter Starting of Electric Drives

Answer 1

To calculate the back EMF (E_b) developed in the motor when it operates on 34\frac{3}{4}43​ of the full load, we can use the following formula for a separately excited DC motor:

Eb=V−IaRaE_b = V - I_a R_aEb​=V−Ia​Ra​

Where:

• EbE_bEb​ is the back EMF,
• $V$ is the supply voltage,
• IaI_aIa​ is the armature current,
• RaR_aRa​ is the armature resistance.

Step 1: Calculate the full-load armature current

The full-load current is given as 2 A. When the motor operates at 34\frac{3}{4}43​ of the full load, the armature current will be:

Ia=34×2 A=1.5 AI_a = \frac{3}{4} \times 2 \, \text{A} = 1.5 \, \text{A}Ia​=43​×2A=1.5A

Step 2: Calculate the back EMF

Given:

• $V=100\text{V}$,
• Ia=1.5 AI_a = 1.5 \, \text{A}Ia​=1.5A,
• Ra=8 ΩR_a = 8 \, \OmegaRa​=8Ω.

Substitute the values into the formula:

Eb=100 V−(1.5 A×8 Ω)E_b = 100 \, \text{V} - (1.5 \, \text{A} \times 8 \, \Omega)Eb​=100V−(1.5A×8Ω) Eb=100 V−12 VE_b = 100 \, \text{V} - 12 \, \text{V}Eb​=100V−12V Eb=88 VE_b = 88 \, \text{V}Eb​=88V

Thus, the back EMF developed in the motor is 88 V.

The correct answer is (c) 88 V.