Question

Two identical moving coil galvanometers have $10\Omega$ resistance and full scale deflection at $2\mu A$ current. One of them is converted into a voltmeter of $100mV$ full scale reading and the other into an Ammeter of $1mA$ full scale current using appropriate resistors. These are then used to measure the voltage and current in Ohm's law experiment with $R=1000\Omega$ resistor by using an ideal cell. Which of the following statement(s) is/are correct?
This question has multiple correct options.
A. The resistance of the voltmeter will be $100k\Omega$
B. The resistance of the ammeter will be $0.02\Omega$ (round of to second decimal place)
C. If the ideal cell is replaced by a cell having internal resistance of $5\Omega$ then the measured value of R will be more than $1000\Omega$
D. The measure value of R will be $978\Omega$<$R$ <$982\Omega$ .

Answer 1

We are given the resistance and the current at which maximum deflection is seen in two identical moving coil galvanometers. We can find the resistance in the voltmeter and ammeter using appropriate equations and verify Ohm’s law using the Ohm’s law equation. Thus we can find the correct options.

Formula used:
$v={{I}_{g}}\left( {{R}_{g}}+{{R}_{v}} \right)$
$S=\dfrac{{{I}_{g}}{{R}_{g}}}{I-{{I}_{g}}}$
$I=\dfrac{V}{R}$

Complete step by step answer:
In the question we are given two identical moving coil galvanometers. Both the galvanometers have a resistance of $10\Omega$ and have a full scale deflection at $2\mu A$ current.
It is said that one of the galvanometers is converted into a voltmeter and the other into an ammeter. The full scale reading of the voltmeter is $100mV$ and in the ammeter is $1mA$.
This voltmeter and ammeter is then used to measure voltage and current in the Ohm’s law experiment using an ideal cell and a resistor of $R=1000\Omega$.

In the three given figures above, figure 1 shows the galvanometer converted into voltmeter, figure 2 shows the galvanometer converted into ammeter and figure 3 shows the voltmeter and ammeter connected together to verify the Ohm’s law.
Now let us consider the figure 1.
Here the full scale deflection of the voltmeter is given as $100mV$ and also we have the current at which the galvanometer shows full scale deflection as $2\mu A$.
Therefore we can calculate the resistance in the voltmeter as,
${{R}_{V}}=R+G$
$\Rightarrow {{R}_{V}}=\dfrac{100mV}{{{I}_{g}}}$
We have, ${{I}_{g}}=2\mu A$
Therefore we get,
$\Rightarrow {{R}_{V}}=\dfrac{100mV}{2}$
$\Rightarrow {{R}_{V}}=50k\Omega$
Therefore the resistance in the voltmeter is $50k\Omega$.
In the first option given it is said that the resistance in the voltmeter is $100k\Omega$.
Hence we can say that this option is wrong.
Now let us consider the figure 2, i.e. the ammeter.
We have the current passed for full scale deflection as $1mA=1000\mu A$ and we have the current for maximum deflection of the galvanometer as
$2\mu A$.
Therefore the current that passes through the shunt resistor will be, $1000-2=998\mu A$.
We have the equation for shunt resistance as,
$S=\dfrac{{{I}_{g}}{{R}_{g}}}{I-{{I}_{g}}}$
We have,
${{R}_{g}}=10\Omega$
${{I}_{g}}=2\mu A$
Therefore we have,
$\Rightarrow S=\dfrac{2\times 10}{1000-2}$
$\Rightarrow S=0.02\Omega$
Therefore we get the resistance in the ammeter as $0.02\Omega$.
In the second option it is said that the resistance in the ammeter will be $0.02\Omega$.
Hence option B is correct.
Now let us consider the figure 3.
Let ‘I’ be the current and ‘E’ be the emf of the cell in the circuit.
From the figure, we get the total resistance in the circuit as,
${{R}_{total}}={{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}}$.
By ohm's law we can write,
$I=\dfrac{E}{{{R}_{total}}}$
$\Rightarrow I=\dfrac{E}{\left( {{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}} \right)}$
Let ‘x’ be the voltage drop in the resistor.
Then we the resistance will be,
$R'=\dfrac{x}{I}$
By Kirchhoff’s rule, we can write
$E-I{{R}_{A}}-x=0$
From this equation we will get ‘x’ as,
$\Rightarrow x=E-I{{R}_{A}}$
We know that $I=\dfrac{E}{\left( {{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}} \right)}$, therefore
$\Rightarrow x=E-\left( \dfrac{E}{\left( {{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}} \right)}\times {{R}_{A}} \right)$
$\Rightarrow x=\dfrac{E\left( {{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}} \right)-E\times {{R}_{A}}}{\left( {{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}} \right)}$
$\Rightarrow x=\dfrac{E{{R}_{A}}-E\times \dfrac{R{{R}_{V}}}{R+{{R}_{V}}}-E{{R}_{A}}}{\left( {{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}} \right)}$
$\Rightarrow x=\dfrac{E\times \dfrac{R{{R}_{V}}}{R+{{R}_{V}}}}{\left( {{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}} \right)}$
Therefore we will get the measured value of resistance as,
$\Rightarrow R'=\dfrac{\left( \dfrac{E\times \dfrac{R{{R}_{V}}}{R+{{R}_{V}}}}{{{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}}} \right)}{\left( \dfrac{E}{{{R}_{A}}+\dfrac{R{{R}_{V}}}{R+{{R}_{V}}}} \right)}$
$\Rightarrow R'=\dfrac{E\times \dfrac{R{{R}_{V}}}{R+{{R}_{V}}}}{E}$
$\Rightarrow R'=\dfrac{R{{R}_{V}}}{R+{{R}_{V}}}$
We have $R=1000\Omega$, ${{R}_{V}}=50k\Omega =50\times {{10}^{3}}\Omega$
Therefore we get,
$\Rightarrow R'=\dfrac{1000\times 50\times {{10}^{3}}}{1000+\left( 50\times {{10}^{3}} \right)}$
$\Rightarrow R'=\dfrac{5\times {{10}^{7}}}{51000}$
$\Rightarrow R'=980.4\Omega$
In the fourth option it is said that the measured value of R will be $978\Omega$<$R$ <$982\Omega$.
We know that $978\Omega$<$980.4\Omega$ <$982\Omega$ .
Therefore we know that option D is correct.
Hence the correct answers are option B and option D.

Note:
In the third option it is said that the ideal cell is replaced by an internal resistance of $5\Omega$.
Let this internal resistance be ‘r’.
Then from figure 3, we can say that the ammeter and the cell are in series.
Hence the total resistance will be, ${{R}_{A}}+r$.
Our calculated expression for the measured resistance is,
$\Rightarrow R'=\dfrac{R{{R}_{V}}}{R+{{R}_{V}}}$
We can see that this equation is independent of the resistance of the ammeter.
Therefore we can say that, even if we replace the ideal cell with an internal resistance the measured value of resistance will not change, i.e.
$R'=980.4\Omega$
And we know that this value is less than $1000\Omega$.
Hence option C is incorrect.