Question

Two identical moving coil galvanometers have 10 H 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 m A 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 resistance of the Voltmeter will be $100k\Omega$
• B. The resistance of the Ammeter will be $0.02\Omega$ (round off to $2^{nd}$ decimal place)
• C. The measured value of $R$ will be $978\Omega <R<982\Omega$
• D. If the ideal cell is replaced by a cell having internal resistance of $5\Omega$ then the measured value of I? will be more than $1000\Omega$

Answer 1

Answer: (B), (C)

Maximum coil current $I_c=2\times 10^{-6}A,R_c=10\Omega$
$\therefore V_{max}$across coil $=2\times 10^{-6}\times 10=2\times 10^{-5}$ volt.
when converted into voltmeter of range 100 mv.
$\left(10+R_0\right)\times 2\times 10^{-6}=100\times 10^{-3}$
$\Rightarrow R_0=49990\Omega$
And when converted into ammeter of range 1 mA, then
$R_s\times 9.98\times 10^{-4}=2\times 10^{-5}\Omega$
$\Rightarrow R_s=0.02\Omega$
Now,
$\therefore R_{eq}=980.48\Omega \therefore I=\frac{V_0}{980.48}and{V}^{\prime }=\frac{V_0\times 50000}{980.48\times 51}$
$R_{\left(measured\right)}=\frac{V^{\prime }}{I}=\frac{50000}{51}=980.4\Omega$