JEE Physics/ Lakshmikanta Satapathy/ Electromagnetism QA part 7/ Question on doubling the range of an ammeter by shunting solved with the related concepts

1. Physics Helpline
L K Satapathy
Electromagnetism 7
Galvanometer & Ammeter
S
G
2I I/2
I/2
X
I

2. Physics Helpline
L K Satapathy
Electromagnetism 7
Question : An ammeter is obtained by shunting a 30 galvanometer with a 30
resistance . What additional resistance should be connected across it to double its
range?
(a) 10 (b) 15 (c) 30 (d) 60
Answer :
2
G S
I
I I  
Let current rating of ammeter = I
S
G
I I/2
I/2
Resistance of Galvanometer G = 30
& Resistance of Shunt S = 30
 Current is equally shared by G & S
Case (i) :

3. Physics Helpline
L K Satapathy
Electromagnetism 7
Correct option = (b)
Current rating of ammeter = 2I
The resistances G , S & X are in parallel
 Current through X = I
Case (ii) :
S
G
2I I/2
I/2
X
IGalvanometer Current = I/2
Shunt Current = I/2
 Voltages across all the resistances are equal
 Current X is double  Resistance X = ½ G = ½ S
 Resistance X = 15  [Ans]

4. Physics Helpline
L K Satapathy
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