2. Solved Papers
  3. VIT Engineering

VIT Engineering VIT Engineering Solved Paper-2010

  • question_answer
    The ratio of the amounts of heat developed in the four arms of a balance Wheatstone bridge, when the arms have resistances \[P=100\,\Omega \], \[Q=10\,\Omega \], \[R=300\,\Omega \] and \[S=30\,\Omega \] respectively is

    A)  3 : 30 : 1 : 10   

    B)  30 : 3 : 10 : 1

    C)  30 : 10 : 1 : 3   

    D)  30 : 1 : 3 : 10  

    Correct Answer: B

    Solution :

    Let i be the total current passing through balanced Wheatstone bridge. Current through arms of resistances P and Q in series is \[{{i}_{1}}=\frac{i\times 330}{330+110}\] \[=\frac{3}{4}i\] and current through arms of resistances R and S in series is \[{{i}_{1}}=\frac{i\times 110}{330+110}=\frac{1}{4}i\] \[\therefore \]  Ratio of heat developed  per sec \[{{H}_{P}}:{{H}_{Q}}:{{H}_{R}}:{{H}_{S}}\] \[={{\left( \frac{3}{4}i \right)}^{2}}\times 100:{{\left( \frac{3}{4}i \right)}^{2}}\times 10:{{\left( \frac{1}{4}i \right)}^{2}}\times 300\] \[:{{\left( \frac{1}{4}i \right)}^{2}}\times 30\] \[=30:3:10:1\]


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