Question

The armature of a 4-pole lap wound dc machine has core length 30 cm, diameter 40 cm, total conductors 500, speed 1200 rpm and current 20 A. For an average flux density of 0.5 T, mechanical power developed is _____W.

• A. 9424

Answer 1

The correct option is A 9424
Here, $I=0.3m,r=0.2mN=1200rpm$

As we know,

$\text{Flux per pole,}\varphi =\frac{2B_m}{\pi }\times \frac{2\pi rl}{P}Wb/pole$

$(\therefore \frac{2B_m}{\pi }=$Average flux density $\left(B_{avg}\right))$

$\varphi =\frac{B_{avg}\times 2\pi rl}{P}$

$\therefore Fluxperpole,\varphi =\frac{0.5\times 2\pi \times 0.2\times 0.3}{4}=0.04712Wb$

General emf,

$E_g=\frac{\varphi ZNP}{60A}=\frac{0.04712\times 500\times 1200\times 4}{60\times 4}=471.2V$

Gross mechanical power developed,

$=E_a{I}_a=471.2\times 20=9424W$