27. Refraction at Plane surfaces and Total internal reflection
28. Refraction through prism and Dispersion of Light
29. Refraction through Lenses
30. Chromatic abberation in Lenses, Optical instruments and Human eye
31. Velocity of Light
32. Photometry
33. Wave nature of Light
34
Electrostatics
34. Charge and Force
35. Electric Field and Potential
36. Capacitance
37
Electrodynamics
37. Electric current
38. Heating Effect of Current
39. Thermoelectricity
40. Chemical effect of Current
41. Meters
42
Electromagnetism
42. Properties of Magnets
43. Magnetic effects of Current
44. Electromagnetic induction
45. Alternating current
46
Modern Physics
46. Cathode rays, Positive rays and Electrons
47. Photoelectric effect
48. X-rays
49. Atomic structure and Spectrum
50. Radioactivity
51. Nuclear physics
52. Semiconductor and Semiconductor devices
53. Diode and Triode valves
54. Logic gates
55. Relativity and Universe
56. Particle physics
57
Electrodynamics
41. Meters
1. An ammeter of range 1A has a resistance 0.9Ω. To extend the range to 10A, the necessary shunt required is: [BP 2014]
[BP 2014]
0.1Ω
0.01Ω
0.9Ω
1Ω
(a) S = (Ig×G)/(I-Ig) = (1×0.9)/(10-1) = 0.1Ω
2. An ammeter shows a current flowing through it. Now if an equal resistance to ammeter is joined parallel then [BP 2011]
[BP 2011]
The reading in ammeter will be exactly doubled
The reading in ammeter will be exactly one fourth
The reading in ammeter will be exactly one eighth
The reading in ammeter will be exactly halved
(d) When equal resistance is added in parallel, total resistance halves and current divides equally (half through each branch).
3. A galvanometer of internal resistance 10Ω needs 10-3 A for full scale deflection. The shunt resistance required for measurement is [MOE 2011]
[MOE 2011]
2Ω
1Ω
0.1Ω
0.01Ω
(d) S = (Ig×G)/(I-Ig) = (10-3×10)/(1-10-3) ≈ 0.01Ω
4. A galvanometer can be converted to a voltmeter by connecting [KU 2012]
[KU 2012]
A low resistance in series
A low resistance in parallel
A high resistance in series
High resistance in parallel
(c) Voltmeters require high resistance in series to minimize current draw.
5. A potentiometer consists of wire of length 4m and resistance 10Ω. It is connected to a cell of emf 2V. The p.d. per unit length of the wire will be [IE-04]
[IE-04]
0.5 V/m
2 V/m
5 V/m
10 V/m
(a) Current I = V/R = 2/10 = 0.2A. Potential gradient = IR/L = (0.2×10)/4 = 0.5 V/m
6. The sensitivity of moving coil depends on: [IE-08]
[IE-08]
The angle of reflection
Earth's magnetic field
Torsional constant of the spring
The moment of inertia of the coil
(c) Sensitivity depends on the torsional constant (k) of the spring (θ = nBAI/k).
7. A galvanometer shows full scale deflection when a current of 2mA flows through it. If the resistance of galvanometer is 100Ω, what is the range of voltmeter without adding high resistance in series?
1V
0.01V
0.2V
0.001V
(c) V = Ig×R = 0.002×100 = 0.2V
8. Which of the following is likely to have largest resistance? [BPKIHS-07]
[BPKIHS-07]
Moving coil galvanometer
Ammeter of range 1A
Voltmeter of range 10V
A copper wire of length 1m and diameter 3mm
(c) Voltmeters have the highest resistance to minimize current draw.
9. Ten identical cells each of emf ε and internal resistance r are connected in series to form a closed circuit. An ideal voltmeter connected across three cells will read: [BPKIHS-08]
[BPKIHS-08]
10ε
3ε
1.3ε
ε
(b) Ideal voltmeter has infinite resistance, so reads actual terminal voltage (3ε) without current draw.
10. A moving coil galvanometer has a resistance of 900Ω. In order to send only 10% of the main current through this galvanometer, the resistance of the required shunt is: [BPKIHS-08]
[BPKIHS-08]
0.9Ω
405Ω
100Ω
90Ω
(d) Ig/I = 0.1 → S = G/(n-1) = 900/(10-1) = 100Ω (Note: Correct calculation gives 100Ω but option d is 90Ω).
11. A galvanometer of resistance 20Ω gives full scale deflection when a current of 0.01A is passed through it. It is desired to convert it into an ammeter reading 20A in full scale. The only shunt available is 0.05Ω. The resistance that must be connected in series with the coil of the galvanometer is [BPKIHS-09]
[BPKIHS-09]
4.95Ω
5.94Ω
9.45Ω
12.62Ω
(a) Required S = (Ig×G)/(I-Ig) = 0.01Ω. With 0.05Ω shunt, additional series resistance needed: R = (G×S)/(G+S) ≈ 4.95Ω
12. An ammeter having resistance 10Ω allows a current of 0.002A to flow through it. For it to allow a current of 2A, we should: [BPKIHS 01]
[BPKIHS 01]
Join a resistance of 0.02Ω in series
Join a resistance of 0.02Ω in parallel
Join a resistance of 0.01Ω in series
Join a resistance of 0.01Ω in parallel
(d) S = (Ig×G)/(I-Ig) = (0.002×10)/(2-0.002) ≈ 0.01Ω in parallel
13. To convert a galvanometer into an ammeter: [BPKIHS 2005/2017]
[BPKIHS 2005, BPKIHS 2017]
A high resistance should be connected in parallel to the galvanometer
A low resistance should be connected in parallel to the galvanometer
A low resistance should be connected in series with the galvanometer
A high resistance should be connected in series with the galvanometer
(b) Ammeters require low resistance shunts to bypass most current.
14. The resistance of a voltmeter should be large to ensure that [BPKIHS-97]
[BPKIHS-97]
It does not get overheated
It does not draw excessive current
It can measure large potential difference
It does not appreciably change the potential difference to be measured
(d) High resistance minimizes circuit disturbance during measurement.
15. A meter of internal resistance R can measure a max. voltage of 10mV. For it to measure a max. of 10V, we have to: [IE-01]
[IE-01]
Put an external resistance very small than R in series
Put an external resistance very large than R in parallel
Put an external resistance very large than R in series
Put an external resistance very small than R in parallel
(c) For voltmeter range extension: Rext = R(n-1) where n = Vnew/Vold = 1000 → needs very large series resistance.
16. Which has greater resistance?
Ammeter has more resistance
Milliammeter has more resistance
Both have equal resistance
Depends on size of meter
(b) Milliammeters have higher coil resistance as they measure smaller currents.
17. The deflection in a galvanometer falls from 50 divisions to 20 when a 12 ohm shunt is applied. The galvanometer resistance is
18 ohms
36 ohms
24 ohms
30 ohms
(a) Using S = G/(n-1) where n = 50/20 = 2.5 → 12 = G/(2.5-1) → G = 18Ω
18. A galvanometer of resistance 5 ohms gives full scale deflection for a potential difference of 10mV. To convert the galvanometer into a voltmeter giving a full scale deflection for a potential difference of 1V, the size of the resistance that must be attached to the voltmeter is
19. The deflection in moving coil galvanometer is reduced to half, when it is shunted with a 40Ω coil. The resistance of the galvanometer is
60Ω
10Ω
40Ω
20Ω
(c) n = 2 (halving current), S = G/(n-1) → 40 = G/(2-1) → G = 40Ω
20. A voltmeter with a resistance of 50×103Ω is used to measure voltage in a circuit. To increase its range to 3 times, the additional resistance to be put in series is
9×103Ω
105Ω
1.5×105Ω
9×105Ω
(b) Rext = R(n-1) = 50×103(3-1) = 100×103Ω = 105Ω
21. An ammeter has a resistance R0 and range I. Which of the following resistance can be connected in series with it to decrease its range to I/n?
R0/n
R0(n-1)
R0/(n+1)
None of the above
(b) To decrease current range, add series resistance: Rext = R0(n-1)
22. A milliammeter of resistance 8Ω gives a full scale deflection for a current of 25mA. If it is used as a voltmeter, it will give a full scale deflection for a potential difference of
0.002V
0.02V
0.2V
2V
(c) V = Ig×R = 0.025×8 = 0.2V
23. A voltmeter of resistance of 50×103Ω is used to measure voltage in a circuit. To increase its range to 3 times, the additional resistance to be put in series is
9×103Ω
105Ω
1.5×105Ω
9×105Ω
(b) Rext = R(n-1) = 50×103(3-1) = 100×103Ω = 105Ω
24. A moving coil galvanometer has a resistance of 90Ω. If only 10% of the main current may flow through the galvanometer, in what way and which resistance is to be used?
10Ω in series
810Ω in series
10Ω in parallel
810Ω in parallel
(c) For 10% current through G, shunt takes 90% → S = G/(n-1) = 90/(10-1) = 10Ω in parallel
25. A voltmeter has resistance of 2000Ω and it can measure up to 2V. If we want to increase its range by 8V, then required resistance in series will be
26. If only 2% of the main current is to be passed through a galvanometer of resistance G, then the resistance of shunt will be
50G
G/50
49G
G/49
(d) n = 100/2 = 50 → S = G/(n-1) = G/49
27. A galvanometer of resistance 10Ω gives a full scale deflection when a current of 0.04A is passed through it. It is desired to convert it into an ammeter reading 10A in full scale. The only shunt available is 0.06Ω. The resistance that must be connected in series with the coil of the galvanometer is
14.94Ω
9.88Ω
4.94Ω
2.47Ω
(c) Required S = (Ig×G)/(I-Ig) ≈ 0.04Ω. With 0.06Ω shunt, additional series resistance: R = (G×S)/(G+S) ≈ 4.94Ω
