JEE Physics 2023 Question with Solution

Given: Full scale deflection current of galvanometer is $I_g = 1\,mA = 10^{-3}\,A$ and galvanometer resistance is $G = 54\,\Omega$.

Find: Which statements are correct for designing a voltmeter of range $50\,V$ and an ammeter of range $10\,mA$.

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

$$V = I_g(G + R)$$

Substituting the given values,

$$50 = 10^{-3}(54 + R)$$ $$54 + R = 50 \times 10^{3}$$ $$R \approx 49946\,\Omega \approx 50\,k\Omega$$

Thus, statement (A) is correct.

For an ammeter, a shunt resistance $r$ is connected in parallel with the galvanometer. Total current is $I = 10\,mA$ and galvanometer current is $I_g = 1\,mA$. So, current through shunt is

$$I_s = I - I_g = 9\,mA$$

Using the shunt formula,

$$r = \frac{I_g}{I_s} \, G = \frac{1}{9} \times 54 = 6\,\Omega$$

Thus, statement (C) is correct.

Therefore, the correct statements are (A) and (C). The correct option is A.

Given: The galvanometer has $I_g = 1\,mA$ and $G = 54\,\Omega$.

Find: The correct pair of statements.

A galvanometer is converted into a voltmeter by adding a large resistance in series, so the series resistance must be close to

$$\frac{V}{I_g} = \frac{50}{10^{-3}} = 50\,k\Omega$$

Since $54\,\Omega$ is very small compared to this value, the required series resistance is approximately $50\,k\Omega$. So statement (A) is valid.

For ammeter conversion, the shunt must carry the remaining current. Here, the shunt carries $9\,mA$ while the galvanometer carries $1\,mA$, so the shunt resistance must be smaller than $54\,\Omega$ by a factor of $9$:

$$r = \frac{54}{9} = 6\,\Omega$$

So statement (C) is valid.

Therefore, the correct option is A.