A galvanometer has a resistance of `3663Omega`. A shunt `S` is connected across it such that (`1//34`) of the total current passes through the galvanometer. Then the value of the shunt is

Options

A. `222Omega`
B. `111Omega`
C. `11Omega`
D. `22Omega`

Explanation

To find the value of the shunt resistance \( S \) connected across a galvanometer with a resistance of \( R_g = 3663 \, \Omega \), we can follow these steps:

### Step 1: Understand the current distribution
Given that \( \frac{1}{34} \) of the total current \( I \) passes through the galvanometer, the current through the galvanometer \( I_g \) can be expressed as:
\[ I_g = \frac{I}{34} \]
The current through the shunt \( I_s \) will then be:
\[ I_s = I - I_g = I - \frac{I}{34} = \frac{34I}{34} - \frac{I}{34} = \frac{33I}{34} \]

### Step 2: Write the potential difference equations
The potential difference across the galvanometer \( V_g \) can be expressed as:
\[ V_g = I_g \cdot R_g = \left(\frac{I}{34}\right) \cdot 3663 \]
The potential difference across the shunt \( V_s \) can be expressed as:
\[ V_s = I_s \cdot S = \left(\frac{33I}{34}\right) \cdot S \]

### Step 3: Set the potential differences equal
Since the shunt and the galvanometer are connected in parallel, the potential differences across both must be equal:
\[ V_g = V_s \]
Substituting the expressions we derived:
\[ \left(\frac{I}{34}\right) \cdot 3663 = \left(\frac{33I}{34}\right) \cdot S \]

### Step 4: Cancel out \( I \) and simplify
We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ \frac{3663}{34} = \frac{33S}{34} \]
Multiplying both sides by \( 34 \):
\[ 3663 = 33S \]

### Step 5: Solve for \( S \)
Now, divide both sides by \( 33 \):
\[ S = \frac{3663}{33} \]
Calculating the right side:
\[ S = 111 \, \Omega \]

### Final Answer
The value of the shunt resistance \( S \) is \( 111 \, \Omega \).

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