Two identical moving coil galvanometer have 10Ω resistance and full scale deflection at 2 current. One of them is converted into a voltmeter of 100mV full scale reading and the other into an Ammeter of 1mA full scale

Question

Quantrex Academy · Accepted Answer

Correct answer: B. The measured value of R will be 978Ω < R < 982Ω .

Two identical moving coil galvanometer have 10Ω resistance and full scale deflection at 2 current. One of them is

Voltmeter rom galvanometer arrangement.
Given:

{R}_{g} = 10Ω

and

{I}_{g} = 2\muA

Required voltage range,

V = 100 \times {10}^{- 3}volts

⇒ V = {I}_{g}\left({R}_{g} + {R}_{v}\right) = {10}^{- 1}

⇒ \dfrac{{10}^{- 1}}{2 \times {10}^{- 6}} = {R}_{g} + {R}_{v}\left(as, {I}_{g} = 2\muA\right)

⇒ 5 \times {10}^{4}Ω \approx {R}_{v} ( {R}_{v} < {10}^{5}Ω ) ⇒

Option

( A )

is incorrect.
Two identical moving coil galvanometer have 10Ω resistance and full scale deflection at 2 current. One of them is

Galvanometer to Ammeter arrangement

\Delta{V}_{G} = \Delta{V}_{S}

{I}_{g} ⋅ {R}_{g} = \left(I - {I}_{g}\right) ⋅ S

⇒ S = \dfrac{{I}_{g} ⋅ {R}_{g}}{I - {I}_{g}} = \dfrac{2 \times {10}^{- 6} \times 10}{{10}^{- 3} - 2 \times {10}^{- 6}}

⇒ S = 2 \times {10}^{- 5} \times {10}^{3} = 2 \times {10}^{- 2}

⇒ S = 20mΩ ⇒ ( C )

is correct.
Circuit diagram for ohm's law
Two identical moving coil galvanometer have 10Ω resistance and full scale deflection at 2 current. One of them is

Resistance ammeter,

{R}_{A} = \dfrac{20 \times {10}^{- 3} \times 10}{10} = 20 \times {10}^{- 3}Ω

i = \dfrac{ε}{\left(\dfrac{1000 \times 50 \times {10}^{3}}{51 \times {10}^{3}}\right)} = \dfrac{51ε}{5 \times {10}^{4}}\left(∵ {R}_{A} \to 0\right)

{i}^{′} = i\left(\dfrac{{R}_{v}}{51 \times {10}^{3}}\right) = \dfrac{ε}{1000}

Measured resistance

∴ {R}_{m} = \dfrac{{i}^{′} \times 1000}{i} = \dfrac{ε}{51ε} \times 5 \times {10}^{4} = \dfrac{5 \times {10}^{4}}{51} = 980.4Ω

If the voltmeter shows full scale deflection then

\dfrac{ε}{980} \times \left(\dfrac{1000}{51 \times {10}^{3}}\right) \times 5 \times {10}^{4} = {10}^{- 1}

∴ ε = 999.6mv

Since

{i}_{A} = 1mA

so max reading of R can be

\dfrac{999.6mV}{1mA} = 999.6Ω

Two identical moving coil galvanometer have 10Ω resistance and full...

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