A galvanometer of resistance 50 Ω Ω is connected to a battery of 3 V along with a resistance of 2950 Ω 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
Text Solution
Verified by ExpertsThe correct answer is:
D
Current through the galvanometer
I = 
= 10 –3 A
Current for 30 divisions = 10 –3 A
30 = 10 –3 A
20 Current for 20 divisions = 
×20
= 
× 10 –3 A
For the same deflection to obtain for 20 division, let resistance added be R
∴ 
R = 4450 Ω Ω
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The current through the ammeter shown in figure is 1 A. If each of the 4 Ω Ω resistor is replaced b…
The meter-bridge wire AB shown in figure is 50 cm long. When AD = 30 cm, no deflection occurs in th…
The potential gradient of a potentiometer wire is defined as
The unit of potential gradient is
The length of the potentiometer wire is kept larger so that the value of potential gradient may
For the same potential difference, a potentiometer wire is replaced by another one of a high specif…