JEE Advanced · Physics · 20. Magnetism & Current

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?

A The resistance of the Voltmeter will be $100 k Ω .$
B The measured value of $R$ will be $978 Ω < R < 982 Ω .$
C The resistance of the Ammeter will be $0.02 Ω$ (round off to $2^{n d}$ decimal place)
D 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 $1000 Ω .$

Verified Solution

Answer & Solution

Correct Answer

(C) The resistance of the Ammeter will be $0.02 Ω$ (round off to $2^{n d}$ decimal place)

Step-by-step Solution

Detailed explanation

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 $\therefore R_m=\frac{i \times 1000}{i}=\frac{\varepsilon}{51 \varepsilon} \times 5 \times 10^4=\frac{5 \times 10^4}{51}$$=980.4 \Omega$
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

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From JEE Advanced

Explore more questions on app