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BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    The potential difference in volt across the resistance R., in the circuit shown in figure, is \[({{R}_{1}}=15\,\Omega ,\,{{R}_{2}}=15\,\Omega ,\,{{R}_{3}}=30\,\Omega ,\,{{R}_{4}}=35\,\Omega )\]

    A) 5                                            

    B) 7.5

    C) 15                         

    D) 12.5

    Correct Answer: C

    Solution :

    Total resistance of the circuit                       \[R+\frac{\left( {{R}_{1}}+{{R}_{2}} \right)\times {{R}_{3}}}{({{R}_{1}}+{{R}_{2}})+{{R}_{3}}}\] \[35+\frac{(15+15)\times 30}{(15+15)+30}\] \[=35+\frac{30\times 30}{30+30}=50\,\Omega \] Current in circuit, \[i=\frac{50}{50}=1A\] Current through \[{{R}_{3}}i={{i}_{2}}=\frac{1}{2}A\] Potential difference across \[{{R}_{3}}\] \[=i\times {{R}_{3}}=\frac{1}{2}\times 30=15V\]


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