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

Question

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 (A) 3 A (B) 6 A (C) 2 A (D) 1 A

Tardigrade Admin · Accepted Answer

We know that back emf =V−IR ⇒ back emf =12-2(1) ⇒ back emf =12-2=10 V ⇒ 10 % of this emf =0.1 × 10=1 V ⇒ Hence the current =1 × 1=1 A hence the current drawn by loaded motor is 2+1=3 A

Q. 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

3 A

6 A

2 A

1 A

We know that back emf $= V - I R$
$\Rightarrow$ back emf $= 12 - 2 (1)$
$\Rightarrow$ back emf $= 12 - 2 = 10 V$
$\Rightarrow 10%$ of this emf $= 0.1 \times 10 = 1 V$
$\Rightarrow$ Hence the current $= 1 \times 1 = 1 A$
hence the current drawn by loaded motor is $2 + 1 = 3 A$

Q. A simple electric motor has an armature resistance of 1Ω and runs from a d.c. source of 12V. It draws a current of 2A 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

3 A

6 A

2 A

1 A

We know that back emf =V−IR ⇒ back emf =12−2(1) ⇒ back emf =12−2=10V ⇒10% of this emf =0.1×10=1V ⇒ Hence the current =1×1=1A hence the current drawn by loaded motor is 2+1=3A

Q. 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

We know that back emf $= V - I R$
$\Rightarrow$ back emf $= 12 - 2 (1)$
$\Rightarrow$ back emf $= 12 - 2 = 10 V$
$\Rightarrow 10%$ of this emf $= 0.1 \times 10 = 1 V$
$\Rightarrow$ Hence the current $= 1 \times 1 = 1 A$
hence the current drawn by loaded motor is $2 + 1 = 3 A$