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BCECE Engineering BCECE Engineering Solved Paper-2015

  • question_answer
    When a battery of emf \[\text{8 V}\] with internal resistance \[0.5\Omega \], is charged by a \[\text{12}0\text{V}\] DC supply using a series resistance of\[15.5\Omega ,\], then the terminal voltage of the battery is

    A) \[\text{11}.\text{5V}\]  

    B) \[\text{2}0\text{V}\]   

    C) \[\text{21}.\text{5 V}\]                                

    D) \[\text{12}.\text{3 V}\]

    Correct Answer: A

    Solution :

    Here, emf of battery = 8V Internal resistance of the battery, \[r=0.5\Omega \]. Voltage of D.C. supply = 120 V External resistance, R = 15.5\[\Omega \], When the battery of emf 8 V is charged from a DC supply of 120 V, the effective emf in the circuit is \[\varepsilon =120-8=112V\] Total resistance of the circuit = R + r \[=(15.50.5)\Omega =16\Omega \] \[\therefore \]Current in the circuit during charging \[I=\frac{\varepsilon }{R+r}=\frac{112}{16}=7A\] \[\therefore \] Terminal voltage of the battery, V = emf of the battery + voltage drop across the battery \[=8+Ir=8+7(0.5)=11.5V\]


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