Question:

In the Wheatstone's network given, \( P = 100 \, \Omega \), \( Q = 200 \, \Omega \), \( R = 150 \, \Omega \), \( S = 300 \, \Omega \), the current passing through the battery (of negligible internal resistance) is

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For a Wheatstone bridge, when the bridge is unbalanced, the current in the circuit can be calculated using Ohm's law and equivalent resistances.
Updated On: Jan 6, 2026
  • \( 0.36 \, \text{A} \)
  • \( 0 \, \text{A} \)
  • \( 0.18 \, \text{A} \)
  • \( 0.72 \, \text{A} \)
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The Correct Option is A

Solution and Explanation

In a Wheatstone bridge, when the bridge is balanced, the current passing through the galvanometer is zero. By solving for the current in this unbalanced configuration, we find that the current through the battery is \( 0.36 \, \text{A} \).

Step 2: Conclusion.
The current passing through the battery is \( 0.36 \, \text{A} \), corresponding to option (a).
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