A galvanometer of resistance 150 ohm is shunted such that only 1/11 of the main current flows through the galvanometer. The resistance of the shunt is, (in ohm)

Options

A. 5
B. 10
C. 15
D. 25

Answer

To find the resistance of the shunt in the given problem, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We have a galvanometer with a resistance of \( R_g = 150 \, \Omega \) and it is shunted such that only \( \frac{1}{11} \) of the main current \( I \) flows through the galvanometer.

2. **Define Variables**:
Let \( I_g \) be the current through the galvanometer. According to the problem, \( I_g = \frac{I}{11} \).

3. **Use the Concept of Parallel Resistance**:
The galvanometer and the shunt resistor \( R_s \) are in parallel. The equivalent resistance \( R_{eq} \) of the parallel combination can be given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R_g} + \frac{1}{R_s} \]

4. **Relate the Currents**:
The total current \( I \) splits into two parts: \( I_g \) through the galvanometer and \( I_s \) through the shunt. We know:
\[ I_s = I - I_g = I - \frac{I}{11} = \frac{10I}{11} \]

5. **Apply Ohm's Law**:
The voltage across both resistors is the same. Therefore, we can write:
\[ I_g R_g = I_s R_s \]
Substituting the values we have:
\[ \left(\frac{I}{11}\right) \cdot 150 = \left(\frac{10I}{11}\right) \cdot R_s \]

6. **Cancel \( I \)**:
Since \( I \) is common in both sides, we can cancel it out (assuming \( I \neq 0 \)):
\[ \frac{150}{11} = \frac{10}{11} R_s \]

7. **Solve for \( R_s \)**:
Multiply both sides by \( 11 \) to eliminate the fraction:
\[ 150 = 10 R_s \]
Now, divide by \( 10 \):
\[ R_s = 15 \, \Omega \]

### Final Answer:
The resistance of the shunt \( R_s \) is \( 15 \, \Omega \).