Question

A 200-V Dc generator supplies 4 kW at a terminal voltage of 200 V, the armature resistance being 0.5 Ω. If the machine is operated as a motor at the same terminal voltage with the same armature current, find the ratio of the generator speed N_g, to the motor speed N_m .
1. \(\dfrac{N_g}{N_m}=1.25\)
2. \(\dfrac{N_g}{N_m}=1.105\)
3. \(\dfrac{N_g}{N_m}=0.905\)
4. \(\dfrac{N_g}{N_m}=0.833\)

Answer 1

Correct Answer - Option 2 : \(\dfrac{N_g}{N_m}=1.105\)

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, the 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:

Given-

V = 200, Ra = 0.5 Ω, Ia = constant, ϕ2 = ϕ1

Let ϕ1 = Generator flux, ϕ2 = Motor flux, Ng = Generator speed, Nm = Motor speed 

∴ \({E_g} = 200 + 20 \times 0.5 = 210\ V\)

\(\begin{array}{l} {E_b} = 200 - 20 \times 0.5 = 190\ V\\ \frac{{{N_m}}}{{{N_g}}} = \frac{{{E_b}}}{{{E_g}}} \times \frac{{{ϕ _1}}}{{{ϕ _2}}} = \frac{{190}}{{210}} = 0.904 \end{array}\)

​\(\dfrac{N_g}{N_m}=\frac{E_g \times \phi_2}{E_b \times \phi_1}\) 

​\(\dfrac{N_g}{N_m}=\frac{210}{190}\)

\(\dfrac{N_g}{N_m}=1.105\)