Answer the following. (a) Briefly describe how the current sensitivity of a moving coil galvanometer can be increased. (b) A galvanometer shows full scale deflection for current I_g. A resistance R_1 is required to convert it into a voltmeter of range (0 – V) and a resistance R_2 to convert it into a voltmeter of range (0 – 2V). Find the resistance of the galvanometer.

PhysicsCBSE Class 12 · Board exam 2023, set 55/5/1 · 3 marks · Short answer

CBSE Class 12 Physics 2023: question 29

Set in the CBSE Class 12 board exam of 2023, paper 55/5/1, section C, question 29, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q29 · 3 marks · Short answerMoving Charges and Magnetism

Answer the following.

(a)
Briefly describe how the current sensitivity of a moving coil galvanometer can be increased.
0 marks
(b)
A galvanometer shows full scale deflection for current $I_{g}$ . A resistance $R_{1}$ is required to convert it into a voltmeter of range (0 – V) and a resistance $R_{2}$ to convert it into a voltmeter of range (0 – 2V). Find the resistance of the galvanometer.
0 marks

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 55/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Recall the formula for current sensitivity in terms of N, B, A and k.

Hint 2

For a voltmeter, the full-scale current $I_{g}$ flows through G and the series resistance together. Write one equation for each range.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(a) Current sensitivity is $I_{s} = \frac{ϕ}{I} = \frac{N B A}{k}$ . It can be increased by:
- increasing the number of turns $N$ ;
- increasing the magnetic field $B$ , using a strong magnet;
- increasing the area $A$ of the coil;
- using a suspension wire of smaller torsion constant $k$ .
1 mark
(b) To convert the galvanometer into a voltmeter, a high resistance is put in series with it, so $V = I_{g} (G + R)$ . For the range $0$ – $V$ : $V = I_{g} (G + R_{1})$ ...(1)
½ mark
For the range $0$ – $2 V$ : $2 V = I_{g} (G + R_{2})$ ...(2)
½ mark
Dividing (2) by (1): $2 = \frac{G + R _{2}}{G + R _{1}}$ , so $2 G + 2 R_{1} = G + R_{2}$ , which gives $G = R_{2} - 2 R_{1}$ .
1 mark

Answer: (a) Increase N, B or A, or decrease k; (b) G = R₂ − 2R₁

Where the marks are

Any correct method to increase current sensitivity, with $I_{s} = N B A / k$ (1 mark)
V = I_g(G + R₁) (½ mark)
2V = I_g(G + R₂) (½ mark)
Solving to get G = R₂ − 2R₁ (1 mark)

Revise the chapter

Chapter: Moving Charges and Magnetism (notes, formulas, where marks are lost and every past question)
Topic: Moving Coil Galvanometer

Practise this chapterBoard-style questions from Moving Charges and Magnetism, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 55/5/1: every question, by section.

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CBSE Class 12 Physics 2023: question 29

Hints

Step-by-step solution

Revise the chapter

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