CET Karnataka Medical CET - Karnataka Medical 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 Wheat Stones 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 \[{{R}_{1}}=P+R\] \[=10+15=25\Omega \] Similarly, Q and S are in .series, so Resistance \[{{R}_{2}}=R+S\] \[=20+30=50\Omega \] 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,     \[I=\frac{V}{R}=\frac{6}{\frac{50}{3}}=0.36A\]


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