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A galvanometer of resistance 50Ω is connected to a battery of 3 V along with a resistance of 2950 Ω in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
Options
(a) 6050 Ω
(b) 4450 Ω
(c) 5050 Ω
(d) 5550 Ω
Correct Answer:
4450 Ω
Explanation:
I=3 / (50+2950) = 10⁻³ A
Current for 30 divisions=10⁻³ A
Current for 20 divisions=10⁻³ * 20/30
=2*10⁻³ / 3 Amperes
For the same deflection to obtain for 20 divisions let resistance added be R, therefore,
2*10⁻³ / 3 = 3 / (50+R)
(50+R) = 9 *10⁻³ / 2
(50+R) = 4500
Therefore R=4450Ω
Related Questions: - If potential(in volts) in a region is expressed as V(x,y,z) = 6xy – y + 2yz,
- The moment of inertia of a semicircular ring about the centre is
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Topics: Alternating Current
(96)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If potential(in volts) in a region is expressed as V(x,y,z) = 6xy – y + 2yz,
- The moment of inertia of a semicircular ring about the centre is
- A galvanometer acting as a voltmeter will have
- Which of the following rods of same material undergoes maximum elongation
- The primary of a transformer when connected to a dc battery of 10 volt draws a current
Topics: Alternating Current (96)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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