CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2007

  • question_answer
    In the Wheatstone's network given, \[P=10\Omega \], \[Q=20\,\Omega ,\,\,R=15\,\Omega ,\,\,S=30\,\Omega \] the current passing through the battery (of negligible internal resistance) is

    A)  0.36 A                                  

    B)  zero

    C)  0.18 A                                  

    D)  0.72 A

    Correct Answer: A

    Solution :

    The balanced condition for Wheatstones bridge is            \[\frac{P}{Q}=\frac{R}{S}\] as is obvious from the given values. No, current flows through galvanometer is zero. Now, P and R are in series, so Resistance \[{{R}_{1}}=P+R\]                                 \[=10+15=250\] Similarly, Q and S are in series, so Resistance \[{{R}_{2}}=R+S\]                                 \[=20+30=500\] Net resistance of the network as \[{{R}_{1}}\] and \[{{R}_{2}}\]are in parallel                 \[\frac{1}{R}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\] \[\therefore \]  \[R=\frac{25\times 50}{25+50}=\frac{50}{3}\Omega \] Hence,  \[I=\frac{V}{R}=\frac{6}{50}=0.36\,A\]


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