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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    A heater A gives out 300 W of heat when connected to a 200 V DC supply. A second heater B gives out 600 W when connected to a 200 V DC, supply. If a series combination of the two heaters is connected to a 200 V DC supply, the heat output will be:

    A)  900 W                  

    B)         450 W  

    C)         300 W                  

    D)         200 W

    E)  100 W                                

    Correct Answer: D

    Solution :

    Let\[{{R}_{1}}\]and\[{{R}_{2}}\]be the resistances, then \[H=\frac{{{V}^{2}}}{R}\] \[\therefore \]  \[{{R}_{1}}=\frac{{{(200)}^{2}}}{300},{{R}_{2}}=\frac{{{(200)}^{2}}}{600}\] In series, \[R={{R}_{1}}+{{R}_{2}}=\frac{{{(200)}^{2}}}{300}+\frac{{{(200)}^{2}}}{600}\] \[=\frac{{{(200)}^{2}}}{100}\left[ \frac{1}{3}+\frac{1}{6} \right]\]                 \[=200\,\Omega \] Heat output, \[H=\frac{{{V}^{2}}}{R}=\frac{{{(200)}^{2}}}{200}=200\,V\]


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