The document discusses a physics problem involving an ammeter created by shunting a 30Ω galvanometer, along with its required additional resistance to double its range. It indicates that the correct answer for the additional resistance needed is 15Ω. The explanation includes the relationships between various currents and resistances in both cases presented.

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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