Question

A grinder motor is designed to operate at a current of \[5\,{{A}}\] and at a potential difference \[200\,{{V}}\] . What resistance must be connected in series with the motor so as to maintain the rated current when it is operated on a \[220\,{{V}}\] line?

Answer 1

Hint: First of all, we will apply the Ohm’s law and find out the resistance of the motor. Then we will again apply the Ohm’s law for the second case, by connecting a new resistor in series with the motor. Magnitude of current will remain the same in series connection. We will manipulate accordingly and find out the result.

Complete step by step answer:
In the given question, we are supplied the following data:
The rated current of the grinder motor is \[5\,{{A}}\] .
The rated potential difference of the grinder motor is \[200\,{{V}}\] .
We are asked to find the resistance which must be connected in series with the motor so as to maintain the rated current when it is operated on a \[220\,{{V}}\] line.

To begin with, first we need to find the resistance present inside the motor, then we have to adjust the new resistor, when the voltage is raised to \[220\,{{V}}\] . But in the second case, the resistor must be connected in series with the motor, so the current flow will remain the same. As we know, that in a series connection the current flow remains the same.

Now, we use the Ohm’s law to find the resistance in the motor.
By Ohm’s law we have,
\[V = iR\] …… (1)
Where,
\[V\] indicates potential difference.
\[i\] indicates the current flow.
\[R\] indicates the resistance of the motor.
By substituting required values in the equation (1), we get:
$
V = iR \\
\Rightarrow 200 = 5 \times R \\
\Rightarrow R = \dfrac{{200}}{5} \\
\Rightarrow R = 40\,{{\Omega }} \\
$
The resistance of the motor is found out to be \[40\,{{\Omega }}\] .
Now, for the second case, we connect a new resistor \[r\] in series with the motor.
So, the net resistance will now be \[\left( {40 + r} \right)\,{{\Omega }}\] .
So, again, we apply the Ohm’s law, we have:
$
V = i\left( {R + r} \right) \\
\Rightarrow 220 = 5 \times \left( {40 + r} \right) \\
\Rightarrow 220 = 200 + 5r \\
\Rightarrow 5r = 20 \\
$
Again, we further simplify,
$
\Rightarrow r = \dfrac{{20}}{5} \\
\Rightarrow r = 4\,{{\Omega }} \\
$
Hence, the resistance must be connected in series with the motor so as to maintain the rated current when it is operated on a \[220\,{{V}}\] line is \[4\,{{\Omega }}\] .

Note: While solving this problem, many students seem to have confusion regarding the magnitude of current in the second case. It should be noted that in series connection, the magnitude of current flow remains the same throughout the circuit. However, in case of parallel connection, the current flowing the different components are different but the potential difference across all the components remains the same.