Question

A galvanometer of resistance $400\,\Omega$ can measure a current of $1\,{\text{mA}}$. To convert it into a voltmeter of range $8\,{\text{V}}$, the required resistance is?
A. $4600\,\Omega$
B. $5600\,\Omega$
C. $6600\,\Omega$
D. $7600\,\Omega$

Answer 1

Use the expression for Ohm’s law. The given resistance is the resistance of the galvanometer. We need to determine the resistance required to achieve desirable range of the voltmeter. To determine this resistance, rewrite Ohm’s law in terms of the galvanometer resistance and the required resistance and solve it.

Formulae used:
The expression for Ohm’s law is given by
$V = IR$ …… (1)
Here, $V$ is the potential difference, $I$ is the current and $R$ is the resistance.

Complete step by step answer:
The resistance of the galvanometer is $400\,\Omega$ and the current measured by the galvanometer is $1\,{\text{mA}}$.
${R_G} = 400\,\Omega$
$I = 1\,{\text{mA}}$
We want the range of the voltmeter to be $8\,{\text{V}}$.
$V = 8\,{\text{V}}$
Convert the unit of the current in the SI system of units.
$I = \left( {1\,{\text{mA}}} \right)\left( {\dfrac{{{{10}^{ - 3}}\,{\text{A}}}}{{1\,{\text{mA}}}}} \right)$
$\Rightarrow I = 1 \times {10^{ - 3}}\,{\text{A}}$
Hence, the current measured by the galvanometer is $1 \times {10^{ - 3}}\,{\text{A}}$.
We can determine the resistance $R$ required to raise the range of voltmeter to $8\,{\text{V}}$.
The required resistance $R$ is the shunt resistance and it should be connected in series with the galvanometer resistance.
Hence, the net resistance in the circuit is the sum of the galvanometer resistance and the shunt resistance $R$.
Rewrite the expression for Ohm’s law for the present situation.
$V = I\left( {R + {R_G}} \right)$
Substitute $8\,{\text{V}}$ for $V$, $1 \times {10^{ - 3}}\,{\text{A}}$ for $I$ and $400\,\Omega$ for ${R_G}$ in the above equation.
$8\,{\text{V}} = \left( {1 \times {{10}^{ - 3}}\,{\text{A}}} \right)\left( {R + 400\,\Omega } \right)$
$\Rightarrow 8 = 0.001R + 0.4$
$\Rightarrow 8 - 0.4 = 0.001R$
$\Rightarrow 7.6 = 0.001R$
$\Rightarrow R = \dfrac{{7.6}}{{0.001}}$
$\Rightarrow R = 7600\,\Omega$

Therefore, the required resistance to be connected in series is $7600\,\Omega$.

So, the correct answer is “Option D”.

Note:
Don’t forget to convert the unit of the current to the SI system of units i.e. from milliampere to ampere as all the units in the formula are in the SI system of units. One may also assume that the range of the voltmeter that we want is $8\,{\text{V}}$, so the voltage could be anything from $0\,{\text{V}}$ to $8\,{\text{V}}$. But we have to determine the resistance for the maximum range of $8\,{\text{V}}$.