Question

A milliamp meter has a resistance of \[10\,\Omega \] and has a range of \[0.25\,{\text{mA}}\]. How will you convert it into a voltmeter to read voltage up to \[25\,{\text{V}}\].

Answer 1

Hint:Use the formula for the equivalent resistance of the two resistors connected in series. Also use the expression for Ohm’s law. First recall how to convert a milliamp meter into a voltmeter. Then calculate the equivalent resistance in the circuit and use Ohm’s law to determine the value of the shunt resistance that should be connected in series in the circuit.

Formulae used:
The equivalent resistance \[{R_{eq}}\] of the two resistors \[{R_1}\] and \[{R_2}\] connected in series is given by
\[{R_{eq}} = {R_1} + {R_2}\] …… (1)
The expression for Ohm’s law is given by
\[V = IR\] …… (2)
Here, \[V\] is the potential difference across the ends of a conductor, \[I\] is the current in the conductor and \[R\] is the resistance of the conductor.

Complete Step by Step Answer:
We have given that the resistance of the milliamp meter is \[10\,\Omega \] and the current range of the milliamp meter is \[0.25\,{\text{mA}}\].
\[R = 10\,\Omega \]
\[ \Rightarrow I = 0.25\,{\text{mA}}\]
We have asked to determine how to convert this milliamp meter into a voltmeter of maximum range \[25\,{\text{V}}\].
\[V = 25\,{\text{V}}\]
Convert the unit of current in the SI system of units.
\[I = \left( {0.25\,{\text{mA}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{A}}}}{{1\,{\text{mA}}}}} \right)\]
\[ \Rightarrow I = 2.5 \times {10^{ - 4}}\,{\text{A}}\]
Hence, the value of the current is \[2.5 \times {10^{ - 4}}\,{\text{A}}\].

To convert the milliamp meter into a voltmeter of required range, a shunt resistor of resistance \[S\] should be converted in series with the resistor \[R\] of milliamp meter.When the shunt resistor is connected in series with the resistor of the milliamp meter, the equivalent resistance of the circuit becomes
\[{R_{eq}} = R + S\]
We can calculate the value of this shunt resistance using Ohm’s law.Rewrite equation (2) for the maximum value of the potential difference measured by the voltmeter.
\[V = I{R_{eq}}\]
Substitute \[R + S\] for \[{R_{eq}}\] in the above equation.
\[V = I\left( {R + S} \right)\]
\[ \Rightarrow S = \dfrac{V}{I} - R\]
Substitute \[25\,{\text{V}}\] for \[V\], \[2.5 \times {10^{ - 4}}\,{\text{A}}\] for \[I\] and \[10\,\Omega \] for \[R\] in the above equation.
\[ \Rightarrow S = \dfrac{{25\,{\text{V}}}}{{2.5 \times {{10}^{ - 4}}\,{\text{A}}}} - \left( {10\,\Omega } \right)\]
\[ \Rightarrow S = 100000 - 10\]
\[ \therefore S = 99990\,\Omega \]

Hence, the milliamp meter can be converted into voltmeter of desired range by connecting a shunt resistance of \[99990\,\Omega \] in series with the resistor in the milliamp meter.

Note:One can also solve the same question by another method. One can also use the formula for the shunt resistance that should be connected in series in the circuit to convert the circuit into voltmeter of desired range. This reduces the efforts of deriving the formula for shunt resistance using formula for equivalent resistance in series arrangement and expression for Ohm’s law.