Question

The shunt required to send 10% of the main current through a moving coil galvanometer of resistance \(99\,\Omega\) is

• A. \(99\,\Omega\)
• B. \(9.9\,\Omega\)
• C. \(9\,\Omega\)
• D. \(10\,\Omega\)
• E. \(11\,\Omega\)

Hint:

Shunt resistance is always small compared to galvanometer resistance.

Answer 1

Answer: (E)

Understanding the Concept: Shunt resistance allows most current to bypass galvanometer.

Step 1: Let total current = \(I\).
Current through galvanometer: \[ I_g = 0.1I \]

Step 2: Current through shunt. \[ I_s = 0.9I \]

Step 3: Equal voltage across parallel branches. \[ I_g R_g = I_s R_s \]

Step 4: Substitute values. \[ 0.1I \times 99 = 0.9I \times R_s \]

Step 5: Solve. \[ 9.9 = 0.9R_s \Rightarrow R_s = 11\,\Omega \]

Step 6: Final Answer. \[ \boxed{11\,\Omega} \]