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 ammeterâs 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
The least count will be given by :
Least count = Scale reading / The number of divisions on the scale
In this case:
Number of divisions = 20
The scale reading = 100 mA
Doing the substitution we have :
Least count = 100/20 = 5 mA
ii) The zero error
At third graduation the reading is :
3 × 5 = 15 mA
Before the key is inserted the pointer is not at 0 and this means there is a zero error.
The zero error is thus :
+15 mA since it is more than zero.
iii) The reading Will be:
At the 15th mark the reading is :
15 × 5 = 75 mA
This is after 300 mA
The current is thus = 300 + 75 = 375 mA
Since there is a zero error the real reading is :
375 - 15 = 360 mA