In `Q.36`, the external resistance that must be connected is series with the main circuit so that the total current in the main circuit remains unaltered even when the galvanometer is shunted is

Options

A. `3663 Omega`
B. `111 Omega`
C. `107.7 Omega`
D. `3555.3 Omega`

Explanation

To solve the problem step by step, we need to determine the external resistance that must be connected in series with the main circuit so that the total current in the main circuit remains unaltered when the galvanometer is shunted.

### Step 1: Understand the Problem
We have a galvanometer with a resistance \( G = 3663 \, \Omega \). A shunt resistance \( S \) is connected across the galvanometer such that \( \frac{1}{34} \) of the total current \( I \) passes through the galvanometer.

### Step 2: Calculate the Shunt Resistance
From the information given, we can express the current through the galvanometer \( I_g \) and the current through the shunt \( I_s \):
- \( I_g = \frac{1}{34} I \)
- \( I_s = I - I_g = I - \frac{1}{34} I = \frac{33}{34} I \)

Using the formula for currents through the galvanometer and shunt, we can write:
\[ \frac{I_g}{I_s} = \frac{G}{S} \]
Substituting the values:
\[ \frac{\frac{1}{34} I}{\frac{33}{34} I} = \frac{G}{S} \]
This simplifies to:
\[ \frac{1}{33} = \frac{G}{S} \]
Thus,
\[ S = 33 G = 33 \times 3663 \, \Omega = 12099 \, \Omega \]

### Step 3: Calculate the Equivalent Resistance of the Shunted Galvanometer
The equivalent resistance \( R_{eq} \) of the galvanometer with the shunt is given by:
\[ R_{eq} = \frac{G \cdot S}{G + S} \]
Substituting the values:
\[ R_{eq} = \frac{3663 \cdot 12099}{3663 + 12099} \]
Calculating the denominator:
\[ 3663 + 12099 = 15762 \]
Now substituting back:
\[ R_{eq} = \frac{3663 \cdot 12099}{15762} \]
Calculating \( R_{eq} \):
\[ R_{eq} \approx 107.7 \, \Omega \]

### Step 4: Find the External Resistance \( R \)
To keep the total current in the main circuit unaltered, the total equivalent resistance of the circuit (including the external resistance \( R \)) must equal the resistance of the galvanometer \( G \):
\[ G = R + R_{eq} \]
Rearranging gives:
\[ R = G - R_{eq} \]
Substituting the values:
\[ R = 3663 - 107.7 \approx 3555.3 \, \Omega \]

### Conclusion
The external resistance that must be connected in series with the main circuit is approximately \( 3555.3 \, \Omega \).