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 $100$ $mV$ full scale reading and the other into an Ammeter of $1$ $mA$ full scale current using appropriate resistors. These are then used to measure the voltage and current in the Ohm's law experiment with $R=1000\Omega$ resistor by using an ideal cell. Which of the following statement(s) is/are correct?

• A. The measured value of $R$ will be $978\Omega <R<982\Omega$
• B. The resistance of the ammeter will be $0.02\Omega$ (round off to $2^{nd}$ 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 resistance of the voltmeter will be $100k\Omega$

Answer 1

Answer: (A), (B)

The resistance of the ammeter will be $0.02\Omega$ (round off to $2^{nd}$ decimal place)

Given, $I_g=2\times {10}^{-6}A{r}_g=10\Omega$

For condition-I,
as a voltmeter $V=i_g(r_g+R_0)$ (where, $R_0$ = shunt resistance)

reading of voltmeter =$100\times {10}^{-3}$
so, $100\times {10}^{-3}=2\times {10}^{-6}(10+R_0)R_0=50k\Omega$
Hence option D is wrong.

For condition-II,
As a ammeter,

resistance of ammeter $R_A=\frac{G\times R_s}{G+R_s}$
$i=i_g\left(\frac{R_s+G}{R_s}\right)i=i_g\left(\frac{r_g}{R_s}\right)$
from the given data,
$1mA=2\times {10}^{-6}\frac{10}{R_s}{R}_s=0.02\Omega$
Hence option B is correct.

For option A,

$\therefore Req=980.48\Omega \therefore I=\frac{V_0}{980.48}$ and $V^{\prime }=\frac{V_0\times 50000}{980.48\times 51}$

$R_{\text{(measured)}}=\frac{V^{\prime }}{I}=\frac{50000}{51}=980.4\Omega$
where, The measured value of $R$ will be $978\Omega <R<982\Omega$.
Hence option A is correct.

For option C,
Internal resistance of cell will not effect for the total resistance measurment.
Hence this option is wrong.