| ⇦ | 
| ⇨ |
A galvanometer coil has a resistance of 15 ohm and gives full scale deflection for a current of 4 mA. To convert it to an ammeter of range 0 to 6 A,
Options
(a) 10 mΩ resistance is to be connected in parallel to the galvanometer
(b) 10 mΩ resistance is to be connected in series with the galvanometer
(c) 0.1 Ω resistance is to be connected in parallel to the galvanometer
(d) 0.1 Ω resistance is to be connected in series with the galvanometer
Correct Answer:
10 mΩ resistance is to be connected in parallel to the galvanometer
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Two springs with spring constants K₁=1500 N/m and K₂=3000 N/m
- The impurity atoms with which pure silicon may be doped to make it a p-type
- The periodic waves of intensities I₁ and I₂ pass through a region at the same time
- Two resistances 3Ω and 6Ω are connected in parallel
- Light of wavelength λ is incident on slit of width d. The resulting diffraction
Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two springs with spring constants K₁=1500 N/m and K₂=3000 N/m
- The impurity atoms with which pure silicon may be doped to make it a p-type
- The periodic waves of intensities I₁ and I₂ pass through a region at the same time
- Two resistances 3Ω and 6Ω are connected in parallel
- Light of wavelength λ is incident on slit of width d. The resulting diffraction
Topics: Alternating Current (96)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Given :-
G=15ohm Ig=4mA =0.004A.
I=6A. A shunt resistor of resistance S is to be connected in parallel with the galvanometer to convert it into an ammeter . The value of S is given by – S = Ig×G / I-Ig.
= 0.004 ×15 / 6-0.004. S= 0.06/5.996 = 0.01ohm =10m ohm.