2. Solved Papers
  3. WB JEE Medical

WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    Three capacitors \[3\mu F,6\mu F\] and \[6\mu F\]are connected in series to a source of 120 V. The potential difference, in volt, across the \[3\mu F\]capacitor will be

    A)  24    

    B)  30     

    C)  40     

    D)  60

    Correct Answer: D

    Solution :

     The combination of three charges in series \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{3}}}\] \[\frac{1}{C}=\frac{1}{3}+\frac{1}{6}+\frac{1}{6}\]\[\Rightarrow \]\[C=\frac{6}{4}=1.5\mu F\] The charge of this circuit \[q=CV=1.5\times 120\] \[q=180\mu C\] The potential difference across the \[3\mu F\] \[q=CV\] \[V=\frac{q}{C}=\frac{180}{3}=60\,V\]


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