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 $100\, k\Omega$
• 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 $I_{c}=2\times10^{-6}A, R_{c}=10\,\Omega$
$\therefore V_{max}$across coil $=2\times10^{-6}\times10=2\times10^{-5}$ volt.
when converted into voltmeter of range 100 mv.
$\left(10 + R_{0}\right) × 2 × 10^{-6} = 100 × 10^{-3}$
$\Rightarrow R_{0}=49990\,\Omega$
And when converted into ammeter of range 1 mA, then
$R_{s} × 9.98 × 10^{-4} = 2 × 10^{-5}\,\Omega$
$\Rightarrow R_{s}=0.02\,\Omega$
Now,
$\therefore R_{eq}=980.48\,\Omega \therefore I=\frac{V_{0}}{980.48} and V'=\frac{V_{0}\times50000}{980.48\times51}$
$R_{\left(measured\right)}=\frac{V'}{I}=\frac{50000}{51}=980.4\,\Omega$