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JEE Main & Advanced Physics Electrostatics & Capacitance Question Bank Grouping of Capacitors

  • 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                    [KCET 1999]

    A)            2, 3, 7                                      

    B)            1.5, 2.5, 8

    C)            1, 5, 6                                      

    D)            4, 2, 6

    Correct Answer: D

    Solution :

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


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