Question

A galvanometer having a resistance of $ 8 \, \Omega $ is shunted by a wire of resistance $ 2 \, \Omega $. If the total current is 1 A, the part of it passing through the shunt will be

• A. 0.25 A
• B. 0.8 A
• C. 0.2 A
• D. 0.5 A

Hint:

For a parallel combination of resistances in a current divider, the current through the branch with lower resistance will be higher.

Answer 1

The Correct Option is B

In this case, we can use the formula for the current division between the galvanometer and the shunt resistance. 
The total current \( I = 1 \, A \) is split between the galvanometer and the shunt. 
We use the concept of parallel resistance to calculate the current through the shunt. 
Let \( I_g \) be the current through the galvanometer, and \( I_s \) be the current through the shunt. 
The total current \( I \) is the sum of the currents through the galvanometer and the shunt: \[ I = I_g + I_s \] 
Using the parallel resistance rule for current division: \[ \frac{I_g}{I_s} = \frac{R_s}{R_g} \] where: 
- \( R_s = 2 \, \Omega \) is the resistance of the shunt, 
- \( R_g = 8 \, \Omega \) is the resistance of the galvanometer. Thus: \[ \frac{I_g}{I_s} = \frac{2}{8} = \frac{1}{4} \] From this ratio, we know that: \[ I_g = \frac{1}{4} \cdot I_s \] Since the total current \( I = 1 \, A \), we have: \[ I_s + I_g = 1 \] \[ I_s + \frac{1}{4} \cdot I_s = 1 \] \[ \frac{5}{4} \cdot I_s = 1 \] \[ I_s = \frac{4}{5} = 0.8 \, \text{A} \] 
Thus, the current passing through the shunt is \( 0.8 \, \text{A} \). 
Therefore, the correct answer is: \[ \text{(2) } 0.8 \, \text{A} \]

Answer 2

To solve the problem of finding the current passing through the shunt, we use the concept that the potential difference across the galvanometer and the shunt is the same because they are connected in parallel.

Given:

• Resistance of the galvanometer, \( R_g = 8 \, \Omega \)
• Resistance of the shunt, \( R_s = 2 \, \Omega \)
• Total current, \( I = 1 \, \text{A} \)

Assuming \( I_g \) is the current through the galvanometer and \( I_s \) is the current through the shunt, the total current is given by:

\[ I = I_g + I_s \]

The potential difference across both the galvanometer and the shunt is:

\[ V_g = I_g \times R_g \]

\[ V_s = I_s \times R_s \]

Since \( V_g = V_s \), we have:

\[ I_g \times R_g = I_s \times R_s \]

Substitute the values of resistances:

\[ I_g \times 8 = I_s \times 2 \]

Simplifying gives:

\[ 4I_g = I_s \]

Using the total current equation:

\[ I = I_g + I_s \]

\[ 1 = I_g + 4I_g \]

\[ 1 = 5I_g \]

Solve for \( I_g \):

\[ I_g = \frac{1}{5} = 0.2 \, \text{A} \]

Now substitute \( I_g \) back into the equation for \( I_s \):

\[ I_s = 4I_g = 4 \times 0.2 = 0.8 \, \text{A} \]

Thus, the current passing through the shunt is 0.8 A.