Question

A moving coil instrument has a resistance $2\,\Omega$ and reads upto $250\,{\text{V}}$ when a resistance $5000\,\Omega$ is connected in series with it. Find the current range of the instrument when it is used as an ammeter with the coil connected across a shunt resistance of 2 milliohm.
A. $50\,{\text{A}}$
B.$25\,{\text{A}}$
C.$100\,{\text{A}}$
D.$200\,{\text{A}}$

Answer 1

Use the formula for equivalent resistance to determine the net resistance through resistors in series. Then determine the current through the series circuit using Ohm’s law. Determine the current through the shunt resistance using the required formula. The current range of the instrument is the sum of current through the main series circuit and through shunt resistance.

Formula used:
The expression for the equivalent resistance ${R_{eq}}$ of the two resistors connected in series is
${R_{eq}} = {R_1} + {R_2}$ …… (1)
Here, ${R_1}$ is the resistance of the first resistor and ${R_2}$ is the resistance of the second resistor.
The expression for Ohm’s law is
$V = IR$ ……. (2)
Here, $V$ is the potential difference across two points, $I$ is the current through the circuit and $R$ is the resistance of the circuit.
The expression for the current through the shunt resistance is given by
${I_{sh}} = \dfrac{{{R_m}{I_m}}}{{{R_{sh}}}}$ …… (3)
Here, ${I_{sh}}$ is the current through the shunt resistance, ${R_{sh}}$ is the shunt resistance, ${I_m}$ is the full scale meter development and ${R_m}$ is the internal resistance.

Complete step by step answer:
The internal resistance ${R_m}$ of $2\,\Omega$ is connected in series with the other resistance $R$ of $5000\,\Omega$.
${R_m} = 2\,\Omega$
$R = 5000\,\Omega$

The potential difference is $250\,{\text{V}}$ and the shunt resistance ${R_{sh}}$ is $2\,{\text{m}}\Omega$.
$V = 250\,{\text{V}}$
${R_{sh}} = 2\,{\text{m}}\Omega$

Determine the equivalent resistance ${R_{eq}}$ of the internal resistance ${R_m}$ and the other resistance $R$.
${R_{eq}} = {R_m} + R$
Substitute $2\,\Omega$ for ${R_m}$ and $5000\,\Omega$ for $R$ in the above equation.
${R_{eq}} = \left( {2\,\Omega } \right) + \left( {5000\,\Omega } \right)$
$\Rightarrow {R_{eq}} = 5002\,\Omega$
Hence, the equivalent resistance is $5002\,\Omega$.
Calculate the current through the two resistors connected in series.
Rearrange Ohm’s law for the current ${I_m}$.
${I_m} = \dfrac{V}{{{R_{eq}}}}$
Substitute $250\,{\text{V}}$ for $V$ and $5002\,\Omega$ for ${R_{eq}}$ in the above equation.
${I_m} = \dfrac{{250\,{\text{V}}}}{{5002\,\Omega }}$
$\Rightarrow {I_m} = 0.04998\,{\text{A}}$
Hence, the current through the series circuit is $0.04998\,{\text{A}}$.
Now determine the current through shunt resistance.
Substitute $2\,\Omega$ for ${R_m}$, $0.04998\,{\text{A}}$ for ${I_m}$ and $2\,{\text{m}}\Omega$ for ${R_{sh}}$ in equation (3).
${I_{sh}} = \dfrac{{\left( {2\,\Omega } \right)\left( {0.04998\,{\text{A}}} \right)}}{{2\,{\text{m}}\Omega }}$
$\Rightarrow {I_{sh}} = \dfrac{{\left( {2\,\Omega } \right)\left( {0.04998\,{\text{A}}} \right)}}{{2 \times {{10}^{ - 3}}\,\Omega }}$
$\Rightarrow {I_{sh}} = 49.98\,{\text{A}}$
Hence, the current through shunt resistance is $49.98\,{\text{A}}$.
The current range $I$ of the instrument is the sum of the current ${I_m}$ through the series circuit and the current ${I_{sh}}$ through the shunt resistance.
$I = {I_m} + {I_{sh}}$
Substitute $0.04998\,{\text{A}}$ for ${I_m}$and $49.98\,{\text{A}}$ for ${I_{sh}}$ in the above equation.
$I = \left( {0.04998\,{\text{A}}} \right) + \left( {49.98\,{\text{A}}} \right)$
$\Rightarrow I = 50.02998\,{\text{A}}$
$\Rightarrow I \approx 50\,{\text{A}}$

Therefore, the current range of the instrument is $50\,{\text{A}}$.

So, the correct answer is “Option A”.

Note:
The current range of any circuit having the shunt resistance connected in it is the sum of the current through the internal resistance or the current through the combination of all the resistors and the current through the shunt resistance.