You visited us 1 times! Enjoying our articles? Unlock Full Access!

Inductors in Circuits

The current d...

Question

The current drawn by a $220 V$ DC series motor of armature resistance $0.50$ and back emf $200 V$ is

40 A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

400 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

450 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

500 A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $40 A$
$E_{b} = V_{t} - I_{a} R_{a}$
where $V_{t}$ = terminal voltage
$E_{b}$ = back emf.
$I_{a}$ = armature current or line current.
$R_{a}$ = Armature resistance.
$⟹ I_{a} = \frac{V_{t} - E_{b}}{R_{a}} = \frac{220 - 200}{0.5} = 40 A$

Suggest Corrections

Similar questions

D C

20 Ω

1.5 A

220 V

D C

200 V

10 A

40

R_{a} = 0.02 Ω

\begin{matrix} Field current (A) & 10 & 20 & 30 & 40 & 60 & 80 Open circuit voltage (V) & 103.5 & 158 & 206 & 230 & 259 & 282 \end{matrix}

Ω

Join BYJU'S Learning Program