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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Mock Test - Capacitance

  • question_answer
    The equivalent capacitance of three capacitors of Capacitance \[{{C}_{1}}:{{C}_{2}}\] and \[{{C}_{3}}\]are connected in parallel is 12 units and product\[{{C}_{1}}.{{C}_{2}}.{{C}_{3}}=48\]. When the capacitors \[{{C}_{1}}\] and \[{{C}_{2}}\] are connected in parallel, the equivalent capacitance is 6 units. Then the capacitance are

    A) 2, 3, 7  

    B) 1.5, 2.5, 8

    C) 1, 5, 6  

    D) 4, 2, 6

    Correct Answer: D

    Solution :

    [d] \[{{C}_{1}}+{{C}_{2}}+{{C}_{3}}=12\]                       ... (i) \[{{C}_{1}}{{C}_{2}}{{C}_{3}}=48\]                         ... (ii) \[{{C}_{1}}+{{C}_{2}}=6\]                            ... (iii) From equations (i) and (iii) \[{{C}_{3}}=6\]                                              ... (iv) From equations (ii) and (iv) \[{{C}_{1}}{{C}_{2}}=8\] Also \[{{({{C}_{1}}-{{C}_{2}})}^{2}}={{({{C}_{1}}+{{C}_{2}})}^{2}}-4{{C}_{1}}{{C}_{2}}\] \[{{({{C}_{1}}-{{C}_{2}})}^{2}}={{(6)}^{2}}-4\times 8=4\] \[\Rightarrow {{C}_{1}}-{{C}_{2}}=2\]                         ... (v) On solving (iii) and (v) \[{{C}_{1}}=4\], \[{{C}_{2}}=2\]


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