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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2000

  • question_answer
    In a typical Wheatstone network the resistances  in  cyclic   order  are  \[A=10\,\Omega ,B=5\Omega ,\,C=4\,\Omega \] and\[D=4\Omega \]for the bridge to be balanced:

    A)  \[10\,\Omega \]should be connected in parallel with A

    B) \[10\,\Omega \] should be connected in series with A

    C) \[5\,\Omega \] should be connected in series with B

    D)  \[5\,\Omega \] should be connected in parallel with B

    Correct Answer: A

    Solution :

    Resistance in upper arms \[A=10\,\Omega ;B=5\Omega \] Resistance in lower arm                                 \[C=4\Omega ;D=4\Omega \] The condition required to form a Wheatstone bridge is                                 \[\frac{A}{B}=\frac{C}{D}\]which does not satisfy this case. If a \[10\,\Omega \] resistance is connected in parallel with A [as per option ] The relation for a new resistance is                                 \[\frac{1}{A'}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}=\frac{1}{5}\] Now the network satisfies the condition of Wheatstone bridge                                 \[\frac{P'}{5}=\frac{R'}{5}or\,\frac{5}{5}=\frac{4}{4}\]


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