Question

A 250 V shunt motor has armature current 100 A and runs at a speed of 300 rpm. The armature resistance is 0.10 Ω. If the shunt field is reduced to 50% of its normal value and the armature current to 50 A, then the new speed of the shunt motor will be:
1. 612.5 rpm
2. 910.5 rpm
3. 312.5 rpm
4. 577.5 rpm

Answer 1

Correct Answer - Option 1 : 612.5 rpm

Concept:

The EMF equation of a DC Machine is

\({E_b} = \frac{{NPϕ Z}}{{60A}}\)

From the above equation,

\(N \propto \frac{{{E_b}}}{ϕ }\)

In a dc generator, induced emf is

Eg = V + IaRa

In a dc motor, back emf is

Eb = V – IaRa

Where,

N is the speed in rpm

ϕ is the flux per pole

P is the number of poles

Z is the number of conductors

V is the terminal voltage

Ia is the armature current

Ra is the armature resistance

Calculation:

Case 1:

Terminal voltage (V_t) = 250 V

Armature current (I_a1) = 100 A

Speed (N₁) = 300 rpm

Armature resistance (R_a) = 0.10 Ω

Back emf, E_b1 = V_t – I_a1R_a

= 250 – 100 × 0.1 = 240 V

Case 2:

Terminal voltage (V_t) = 250 V

Armature current (I_a2) = 50 A

Let the speed, in this case, is N₂

Armature resistance (R_a) = 0.10 Ω

Back emf, E_b2 = V_t – I_a2R_a

= 250 – 50 × 0.1 = 245 V

ϕ₂ = 0.5 ϕ₁

\(\frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{{N_1}}}{{{N_2}}} \times \frac{{{\phi _1}}}{{{\phi _2}}}\)

\( \Rightarrow \frac{{240}}{{245}} = \frac{{300}}{{{N_2}}} \times \frac{{{\phi _1}}}{{0.5{\phi _1}}}\)

⇒ N₂ = 612.5 rpm