Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2007

  • question_answer A capacitor of \[6\mu F\] capacitance is charged up to 100 V. It is touched and then removed with another uncharged capacitor of \[14\mu F,\] then ratio of charges on both and potential at \[6\mu F\] capacitor is

    A)  6/14 and 50V   

    B)  3/7 and 30V

    C)  7/3 and 30V                      

    D)  None of these

    Correct Answer: B

    Solution :

    Given,  \[{{C}_{1}}=6\mu F,\] \[{{C}_{2}}=14\mu F\] \[{{V}_{1}}=100volt\] Charge on capacitor                 \[q={{C}_{1}}{{V}_{1}}=6\times 100=600\mu C\] Redistribution of charge takes place in the ratio of their capacitances. \[\therefore \]  \[\frac{{{q}_{1}}}{{{q}_{2}}}=\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{6}{14}=\frac{3}{7}\] Charge on capacitor \[{{C}_{1}}\] after touching and then \[{{C}_{1}}\]removed,                 \[{{q}_{1}}=\frac{3\times 600}{3+7}=180\mu C\] \[\therefore \] Potential at \[6\mu F\] capacitor

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