Question

A 280 V, separately excited lap wound dc motor has 100 armature conductor. Flux per pole is 0.15 Wb. Armature resistance is 0.2 $\Omega$. If motor is running at 1000 rpm, then armature current is

• A. 175 A
• B. 150 A
• C. 125 A
• D. 200 A

Answer 1

The correct option is B 150 A


Back emf is given by,

$E_b=\frac{NP\varphi Z}{A\times 60}$

$=\frac{1000\times 0.15\times 100}{60}=250V$

$\Rightarrow I_a=\frac{V_t-E_b}{R_a}$

$I_a=\frac{280-250}{0.2}=150A$