Manipal Medical Manipal Medical Solved Paper-2012

  • question_answer
    In the Wheatstones network given,\[P=10\,\Omega ,\] \[Q=20\text{ }\Omega ,\text{ }R=15\text{ }\Omega ,\text{ S}=30\text{ }\Omega ,\]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, \[{{R}_{1}}=P+R\] \[=10+15=25\,\Omega \] 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 \[i=\frac{V}{R}=\frac{0.6}{50}=0.36\,A\]


You need to login to perform this action.
You will be redirected in 3 sec spinner