# 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 s

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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 Ng, to the motor speed Nm .
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. $\dfrac{N_g}{N_m}=0.833$

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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, ϕ= Motor flux, Ng = Generator speed, Nm = Motor speed

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

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

$\dfrac{N_g}{N_m}=1.105$