The range of a milliammeter is 0-500 mA. There are 20 divisions between 0 and 100 mA mark on its scale. During an experiment a student observes ammeter’s pointer at 3rd graduation mark after zero mark when the key is off and the pointer at 15th graduation mark after 300 mA mark when the key is inserted into the plug. Find (i) the least count, (ii) the zero error with proper sign and (iii) the value of current drawn from the battery.
Answers
The answer is (c) 270 mA.
Least count of the ammeter = (100 − 0)mA/20 = 5 mA.
Zero Error = 3 x 5 A = 15 mA
Zero Correction = − 15 mA.
Correct reading of the ammeter = M.S.R. + Division of coincidence x L.C. + Z.C.
= 200 mA + 17 x 5 mA − 15 mA.
= 270 mA
Answer:
Explanation:
Least count = the length between two marks / number of divisions
Least count = 100/20 = 5 mA
When the key is off the reading is :
5 × 3 = 15 mA
This is the zero error.
Since it is after the zero then it is positive.
Zero error = + 15 mA
At the 15th mark the reading is :
15th mark = 5 × 15 = 75 mA
Reading = 300 + 75 = 375 mA
Since the zero error is 15 mA, the correct reading will be :
375 - 15 = 360 mA
This is the correct reading.