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Current Electricity question

2001 · Q170
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Current Electricity question

2001 · Q170

NEETPhysicsCurrent ElectricityMCQ+4 / −1
The resistance of each arm of the Wheatstone's bridge is 10 ohm. A resistance of 10 ohm is connected in series with a galvanometer then the equivalent resistance across the battery will be
  1. A
    10 ohm
  2. B
    15 ohm
  3. C
    20 ohm
  4. D
    40 ohm
View written solutionFree

Correct answer: A

Here, P = Q = S = R = 10 Ω\Omega Ω

since PQ=SR=1{P \over Q} = {S \over R} = 1QP​=RS​=1, so it is balanced Wheatstone bridge and no current pass through galvanometer.

∴\therefore∴ Equivalent resistance is given by

R = (10+10)(10+10)(10+10)+(10+10){{\left( {10 + 10} \right)\left( {10 + 10} \right)} \over {\left( {10 + 10} \right) + \left( {10 + 10} \right)}}(10+10)+(10+10)(10+10)(10+10)​

=20×2040=10Ω= {{20 \times 20} \over {40}} = 10\Omega=4020×20​=10Ω

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