# A separately excited 300 V DC shunt motor under no-load runs at 900 rpm drawing an armature current of 2 A. The armature resistance is 0.5 Ω and the leakage inductance is 0.01 H. When loaded, the armature current is 15 A. Then the speed in rpm is.

A separately excited 300 V DC shunt motor under no-load runs at 900 rpm drawing an armature current of 2 A. The armature resistance is 0.5 Ω and the leakage inductance is 0.01 H. When loaded, the armature current is 15 A. Then the speed in rpm is.

### Right Answer is: 881 RPM

#### SOLUTION

Given Data

Supply Voltage V = 300 V
No-Load Speed No = 900 RPM
Armature current at no-load Ia = 2 A
Armature Resistance Ra = 0.5 Ω
Leakage Inductance = 0.01 H
Armature current at full load Io = 15A

⇒ From the voltage equation, the back EMF of DC motor at no-load

Ebo = V − IaRa

= 300 × 2 × 0.5

Ebo = 299 V

⇒ The back EMF of DC motor on Loaded

Eb = V − IaRa

= 300 × 15 × 0.5

Eb = 292.5 V

Let the on load speed be N. As we know that back EMF of the DC motor is directly proportional to the flux and speed.

Eb ∝ Nφ
Eb ∝ N (φ is constant)

The ratio of emf and speed can be equated at on load and no-load conditions.

$\begin{array}{l}\dfrac{N}{{{N_o}}} = \dfrac{{{E_b}}}{{{E_{bo}}}}\\\\N = \dfrac{{{E_b}}}{{{E_{bo}}}} \times {N_o}\\\\N = \dfrac{{292.5}}{{299}} \times 900\\\\{\text{N = 881 RPM}}\end{array}$

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