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: \[ \text{(2) } 0.8 \, \text{A} \]