Question:

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

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Shunt resistance is always small compared to galvanometer resistance.
Updated On: May 8, 2026
  • \(99\,\Omega\)
  • \(9.9\,\Omega\)
  • \(9\,\Omega\)
  • \(10\,\Omega\)
  • \(11\,\Omega\)
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Solution and Explanation

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:
Conclusion. \[ \boxed{11\,\Omega} \]
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