Question

When a galvanometer is shunted by a resistance 'S', its current capacity increases 'n' times. If the same galvanometer is shunted by another resistance '\(S^1\)', its current capacity will increase to '\(n^1\)'. The value of n in terms of \(n^1\), S and \(S^1\) is

• A. \(\frac{n^1+S}{S^1}\)
• B. \(\frac{S(n^1-1)-S^1}{S}\)
• C. \(\frac{(n^1+1)S^1}{S}\)
• D. \(\frac{S+S^1(n^1-1)}{S}\)

Hint:

Use Ig G = (I - Ig) S for each shunt and equate the galvanometer resistance.

Answer 1

The Correct Option is D

Step 1: Understand the concept
With a shunt \(S\) across a galvanometer of resistance \(G\) and full scale current \(I_g\), the total current \(I\) satisfies \(I_gG = (I - I_g)S\).

Step 2: Use the factor n
Current capacity becomes \(nI_g\), so \(I_gG = (nI_g - I_g)S\), which gives \(G = (n - 1)S\).

Step 3: Second shunt
Similarly, with shunt \(S^1\) we get \(G = (n^1 - 1)S^1\).

Step 4: Equate and solve
\[ (n - 1)S = (n^1 - 1)S^1 \Rightarrow n = 1 + \frac{S^1(n^1 - 1)}{S} = \frac{S + S^1(n^1 - 1)}{S} \]
Option (D).

Final Answer:
n equals (S + S1 (n1 - 1))/S. This is option (D). \[ \boxed{\text{(D) }\frac{S+S^1(n^1-1)}{S}} \]