A simple electric motor has an armature resistance of $1 Ω$ and runs from a d.c. source of $12 V$ . It draws a current of $2 A$ when unloaded. When a certain load is connected to it, its speed reduces by $10$ % of its initial value. The current drawn by the loaded motor is:

6 A

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

3 A

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

2 A

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

1 A

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

Open in App

Solution

The correct option is C $3 A$
We know that $b a c k e m f = V - I R$

$\Rightarrow b a c k e m f = 12 - 2 (1)$

$\Rightarrow b a c k e m f = 12 - 2 = 10 V$

$\Rightarrow 10 % o f t h i s e m f = 0.1 \times 10 = 1 V$

$\Rightarrow H e n c e t h e c u r r e n t = 1 \times 1 = 1 A$

hence the current drawn by loaded motor is $2 + 1 = 3 A$

Suggest Corrections

Similar questions

0.1 Ω

200 Ω

Ω

Join BYJU'S Learning Program