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

Two identical moving coil galvanometer have

10 Ω

resistance and full scale deflection at

2 μ A

current. One of them is converted into a voltmeter of

100 m V

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 Ω

resistor by using an ideal cell. Which of the following statement(s) is/are correct?

The resistance of the Voltmeter will be

100 k Ω .

The measured value of

R

will be

978 Ω < R < 982 Ω .

The resistance of the Ammeter will be

0.02 Ω

(round off to

2^{n d}

decimal place)

If the ideal cell is replaced by a cell having internal resistance of

5 Ω

then the measured value of

R

will be more than

1000 Ω .

Solution:

Voltmeter rom galvanometer arrangement.
Given:

R_{g} = 10 Ω

and

I_{g} = 2 μ A

Required voltage range,

V = 100 \times 10^{- 3} v o l t s

\Rightarrow V = I_{g} (R_{g} + R_{v}) = 10^{- 1}

\Rightarrow \frac{10^{- 1}}{2 \times 10^{- 6}} = R_{g} + R_{v} (a s, I_{g} = 2 μ A)

\Rightarrow 5 \times 10^{4} Ω \approx R_{v} (R_{v} < 10^{5} Ω) \Rightarrow

Option

(A)

is incorrect.

Galvanometer to Ammeter arrangement

Δ V_{G} = Δ V_{S}

I_{g} \cdot R_{g} = (I - I_{g}) \cdot S

\Rightarrow S = \frac{I_{g} \cdot R_{g}}{I - I_{g}} = \frac{2 \times 10^{- 6} \times 10}{10^{- 3} - 2 \times 10^{- 6}}

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

\Rightarrow S = 20 m Ω \Rightarrow (C)

is correct.
Circuit diagram for ohm's law

Resistance ammeter,

R_{A} = \frac{20 \times 10^{- 3} \times 10}{10} = 20 \times 10^{- 3} Ω

i = \frac{ε}{(\frac{1000 \times 50 \times 10^{3}}{51 \times 10^{3}})} = \frac{51 ε}{5 \times 10^{4}} (∵ R_{A} \to 0)

i^{'} = i (\frac{R_{v}}{51 \times 10^{3}}) = \frac{ε}{1000}

Measured resistance

∴ R_{m} = \frac{i^{'} \times 1000}{i} = \frac{ε}{51 ε} \times 5 \times 10^{4} = \frac{5 \times 10^{4}}{51} = 980.4 Ω

If the voltmeter shows full scale deflection then

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

∴ ε = 999.6 m v

Since

i_{A} = 1 m A

so max reading of R can be

\frac{999.6 m V}{1 m A} = 999.6 Ω

Score

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

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