Question

A simple electric motor is having an armature resistance given as $1\Omega $ and runs from a d c source of $12V$. When unloaded, it draws a current of $2A$. If a particular load is connected to it, its speed decreases by $10\text{percent}$ of its initial value. Therefore what will be the current drawn by the loaded motor?
$\begin{align}
  & A.6A \\
 & B.3A \\
 & C.2A \\
 & D.1A \\
\end{align}$

Answer 1

Hint: Back emf can be found by taking the difference of the voltage across the system and the product of current flowing through the system and resistance provided. Find out the $10\text{percent}$ of this emf. Using this find the current flowing. The resultant sum of currents will be the solution. This will help you in answering this question.

Complete step by step answer:
As we all know, the back emf is the difference of the voltage across the system and the product of current flowing through the system and resistance provided. This can be written as an equation,
$\text{back emf=}V-IR$
The voltage has been mentioned as,
$V=12V$
The current through it can be written as,
$I=2A$
Resistance loaded will be,
$R=1\Omega $
Substituting the values in it will give,
$\text{back emf=12}-2\times 1=10V$
The $10\text{percent}$ of the back emf can be shown as,
$\text{10 percent of back emf=}=10\times 0.1=1V$
According to Ohm’s law, the current can be found using the equation which is given as,
$I=\dfrac{V}{R}$
Substituting the values in it,
Hence the current will become,
$I=\dfrac{1}{1}=1A$
Therefore the total current drawn by the loaded motor will be the resultant sum of the current values in the circuit. That is we can write that,
${{I}_{T}}=2+1=3A$
Hence the answer has been calculated. It has been mentioned as option B.

Note:
Ohm's law is a basic law in physics. The law is mentioned as the current which is flowing through a circuit between two positions should be directly proportional to the voltage of the within the points of the circuit. Here the constant of proportionality will be called as resistance.