For designing a voltmeter of range 50 ~V and an ammeter of range 10 ~mA using a galvanometer which has a coil of resistance 54 \Omega showing a full scale deflection for 1 ~mA as in figure. (A) for voltmeter R \approx 50 k \Omega (B) for ammeter r \approx 0.2 \Omega (C) for ammeter r=6 \Omega (D) for voltmeter R \approx 5 k \Omega (E) for voltmeter R \approx 500 \Omega Choose the correct answer from the options given below:

Current Electricity PYQ — JEE Main 2023 · 15 Apr · Shift 1 · Q57 with Solution

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Correct answer: A

We need to design:

For a voltmeter, the galvanometer is connected in series with a high resistance $R$ . At full scale,

$V = I_g(G+R)$

Given $V = 50\,\text{V}$ ,

$50 = 10^{-3}(54 + R)$

$54 + R = 50 \times 10^3 = 50000$

$R = 50000 - 54 = 49946\,\Omega$

Thus,

$R \approx 5.0 \times 10^4\,\Omega = 50\,\text{k}\Omega$

So statement (A) is correct.

Statements (D) and (E) are incorrect.

For an ammeter, a shunt resistance $r$ is connected in parallel with the galvanometer.

Desired ammeter full scale current:

$I = 10\,\text{mA}$

Galvanometer takes only

$I_g = 1\,\text{mA}$

So shunt current is

$I_s = I - I_g = 10 - 1 = 9\,\text{mA}$

Since galvanometer and shunt are in parallel, voltage across them is same:

$I_g G = I_s r$

$10^{-3} \times 54 = 9 \times 10^{-3} \times r$

$54 = 9r$

$r = 6\,\Omega$

So statement (C) is correct, and (B) is incorrect.

Correct statements are:

Hence the correct option is:

$\boxed{\text{A: } (A) \text{ and } (C)}$

Stored correct answer: A

Our derived answer is also A, so they agree.