Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    A 10\[\mu F\] capacitor is charged to a potential difference of 50V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20V. The capacitance of second capacitor is

    A)  10\[\mu F\]                                      

    B)  20\[\mu F\]

    C)  30\[\mu F\]                      

    D)         15\[\mu F\]

    Correct Answer: D

    Solution :

    \[{{C}_{1}}=\cdot 10\mu F,\,\,V=50\,\,V\] Charge on first capacitor                 \[q={{C}_{1}}V\]                 \[=10\times 50=500\mu C\] Common potential\[V\text{=}\frac{\text{total}\,\,\text{charge}}{\text{total}\,\,\text{capacitance}}\] \[\therefore \]  \[20=\frac{500}{10+{{C}_{2}}}\] or            \[20=\frac{500}{10+{{C}_{2}}}\] or            \[{{C}_{2}}=15\mu F\]


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