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: Voltage across the parallel pair
The galvanometer and shunt have the same voltage: $I_gG = I_sS$ with $I_s = I - I_g$.

Step 2: Current capacity
If $I = nI_g$ then $I_s = (n-1)I_g$, so $G = (n-1)S$. Equivalent form: $n = 1 + G/S$.

Step 3: Two shunts
$1 + \dfrac{G}{S} = n$ and $1 + \dfrac{G}{S^1} = n^1$, so $G = S^1(n^1 - 1)$.

Step 4: Combine
$n = 1 + \dfrac{S^1(n^1 - 1)}{S}$, the same as option (D). Check with $S^1 = S$: $n = n^1$, as expected.

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