Question

An electrical meter of internal resistance \[20\,\Omega \] gives a full scale deflection when one milliampere current flows through it. The maximum current that can be measured by using three resistors of resistance \[12\,\Omega \] each, in milliamperes is
A. 10
B. 8
C. 6
D. 4

Answer 1

Hint: We know that, when we connect the resistors parallel to each other, the equivalent resistance decreases. Also, in parallel circuits, the voltage in both arms remains the same. Use Ohm’s law to determine the current in the resistors.

Formula used:
\[I = \dfrac{V}{R}\], where, V is the voltage and R is the resistance of the circuit.

Complete step by step answer:
We know that, when we connect the resistors parallel to each other, the equivalent resistance decreases.

According to Ohm’s law, the current in the circuit is defined as,
\[I = \dfrac{V}{R}\], where, V is the voltage and R is the resistance of the circuit.
From the above expression, we can say that the current in the circuit is inversely proportional to the resistance.
Therefore, the current in the electric meter will be the maximum when the resistance is the minimum. The minimum resistance can be obtained by connecting the three resistors of \[12\,\Omega \] in parallel.
The equivalent resistance of three \[12\,\Omega \] resistors is,
\[\dfrac{1}{{{R_{eq}}}} = \dfrac{1}{{12}} + \dfrac{1}{{12}} + \dfrac{1}{{12}}\]
\[ \Rightarrow \dfrac{1}{{{R_{eq}}}} = \dfrac{3}{{12}} = \dfrac{1}{4}\]
\[ \Rightarrow {R_{eq}} = 4\,\Omega \]

The voltage in the electrical meter when the internal resistance is \[20\,\Omega \] is,
\[V = {I_L}{R_i}\] …… (1)
The voltage across the resistance \[{R_{eq}} = 4\,\Omega \] is,
\[V = {I_2}{R_{eq}}\] …… (2)
Since the voltage in the parallel circuit does not change, we can write,
\[{I_L}{R_i} = {I_2}{R_{eq}}\]
\[ \Rightarrow {I_2} = \dfrac{{{I_L}{R_i}}}{{{R_{eq}}}}\]
Substitute \[1\,mA\] for \[{I_L}\], \[20\,\Omega \] for \[{R_L}\] and \[4\,\Omega \] for \[{R_{eq}}\] in the above equation.
\[{I_2} = \dfrac{{\left( {1\,mA} \right)\left( {20\,\Omega } \right)}}{{4\,\Omega }}\]
\[ \Rightarrow {I_2} = 5\,mA\]
Therefore, the maximum current that can flow through the electric meter is,
\[{I_{\max }} = 1\,mA + 5\,mA\]
\[ \Rightarrow {I_{\max }} = 6\,mA\]

So, the correct answer is “Option C”.

Note:
Students should remember the change in voltage and current in the series and parallel combination. The internal resistance of the electrical resistance is due to the resistance of the electric component in the circuit. This internal resistance is usually neglected in calculations.