A galvanometer having a coil resistance of 100 ohms gives a full scale deflection when a current of one milli-ampere is passed through it. What is the value of resistance which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 amperes? A resistance of the required value is available but it will get burnt if the energy dissipated in it is greater than one watt. Can it be used for the above described conversion of the galvanometer? When this modified galvanometer is connected across the terminals of battery, it shows a current 4 amp. The current drops to 1 amp., when the resistance of 1.5 ohm is connected in series with modified galvanometer. Find the emf and internal resistance of battery.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
S = 
, yes E = 2V, r = 0.5 – 0.01 = 0.49 Ω Ω .
Sol. R g = 100 Ω Ω
I g = 1 mA, S = ?, I = 10 A, P = 1W
100 × 10 –3 = (10 – 10 –3 ) S
S = 
0.01 Ans.
P = (10 – 10 –3 ) 2 × S = (10 – 10 –3 ) × 0.1
10 × 0.1 = 1 Watt
Q Power dissipated is slightly less than 1W so its use is safe Ans.
4 = 
..........(i),
4 = 
1 = 
...........(ii),
+ r = 
From (i) & (ii)
1 = 
, r = 0.49 Ω Ω Ans.
E = 2V Ans.
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