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QJEEAdv19Ph0085: JEE Advanced 2019 Physics - Class 12 JEEAdv19Ph0085 Solution: Two identical moving coil galvanometers have 10 resistance and…

Two identical moving coil galvanometers have 10 Ω\Omega resistance and full scale deflection at 2 μ\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 Ω\Omega resistor by using an ideal cell. Which of the following statement(s) is/are correct?

Multiple Choice Options

  • A The resistance of the voltmeter will be 100 k Ω\Omega .
  • B The resistance of the ammeter will be 0.02 Ω\Omega (round off to 2nd decimal place).
  • C If the ideal cell is replaced by a cell having internal resistance of 5 Ω\Omega then the measured value of R will be more than 1000 Ω\Omega .
  • D The measured value of R will be 978 Ω\Omega < R < 982 Ω\Omega .

Step-by-Step Verified Solution

JEE Advanced 2019 Paper 1 Offline Physics - Current Electricity Question 27 English Explanation 1

V=100×10−3=10−1VV = 100 \times {10^{ - 3}} = {10^{ - 1}}V

V=lg(Rg+Rv)V = {{\mathop{\rm l}\nolimits} _g}({R_g} + {R_v})

10−12×10−6=Rg+Rv{{{{10}^{ - 1}}} \over {2 \times {{10}^{ - 6}}}}={R_g} + {R_v}

5×104Ω≈Rv5 \times {10^4}\Omega \approx {R_v} ( ∵\because {R_v} &lt; {10^5}\Omega )

JEE Advanced 2019 Paper 1 Offline Physics - Current Electricity Question 27 English Explanation 2

IgRg = (I - Ig)S
JEE Advanced 2019 Paper 1 Offline Physics - Current Electricity Question 27 English Explanation 3

S=2×10−6×1010−3−2×10−6S = {{2 \times {{10}^{ - 6}} \times 10} \over {{{10}^{ - 3}} - 2 \times {{10}^{ - 6}}}}

S=2×10−5×103=2×10−2=20mΩS = 2 \times {10^{ - 5}} \times {10^3} = 2 \times {10^{ - 2}} = 20m\Omega

RA=SRgS+Rg=20×10−3×1010+20×10−3{R_A} = {{S{R_g}} \over {S + {R_g}}} = {{20 \times {{10}^{ - 3}} \times 10} \over {10 + 20 \times {{10}^{ - 3}}}}

≈20×10−3Ω\approx 20 \times {10^{ - 3}}\Omega

i=ε(1000×50×10351×103)=51ε5×104i = {\varepsilon \over {\left( {{{1000 \times 50 \times {{10}^3}} \over {51 \times {{10}^3}}}} \right)}} = {{51\varepsilon } \over {5 \times {{10}^4}}}

( ∵\because RA →\to 0)

i′=i(Rv51×103)=ε1000i' = i\left( {{{{R_v}} \over {51 \times {{10}^3}}}} \right) = {\varepsilon \over {1000}}

Measured resistance,

∴\therefore Rm=i′×1000i=ε51ε×5×10=980.4Ω{R_m} = {{i' \times 1000} \over i} = {\varepsilon \over {51\varepsilon }} \times 5 \times 10 = 980.4\Omega

If the voltmeter shows full scale deflection then

ε980×(100051×103)×5×104=10−1{\varepsilon \over {980}} \times \left( {{{1000} \over {51 \times {{10}^3}}}} \right) \times 5 \times {10^4} = {10^{ - 1}}

ε=999.6\varepsilon = 999.6 mV

Since, iA = 1 mA so maximum reading of R can be

999.6 mV1 mA=999.6Ω{{999.6\,mV} \over {1\,mA}} = 999.6\Omega

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