question_answer

Two capacitors of capacitances 3 \[\mu F\]and 6 \[\mu F\]are charged to a potential of 12 V each. They are now connected to each other, with the positive plate to each joined to the negative plate to the other. The potential difference across each will be:

A)  4 V

B)  6 V

C)   zero

D)  3 V

Correct Answer: A

Solution :

\[{{V}_{1}}=12V\,{{V}_{1}}=12V\] \[{{q}_{1}}={{C}_{1}}{{V}_{1}}=36\mu C\] \[{{q}_{2}}={{C}_{2}}{{V}_{1}}=72\mu C\] when the two are connected in series (\[+\upsilon e\] of each to \[-\upsilon e\] of each). \[{{C}_{1}}=3\mu F\] \[\therefore \] \[{{C}_{2}}=9\mu F\] both are charged by 12V battery \[{{C}_{2}}\] \[\therefore \]charge on \[{{C}_{1}}=12\times 3=36\mu C\] charge on \[{{C}_{2}}=12\times 6=72\mu C\] q on\[{{A}_{1}}=+36\mu C\]on\[{{A}_{2}}=-36\mu C\] q on\[{{B}_{1}}=-72\mu C\]on\[{{B}_{2}}=+72\mu C\] Total charge on \[{{A}_{1}}\And {{B}_{1}}=(36-72)\mu C\] \[{{Q}_{1}}=-36\mu C\] Total charge on \[{{A}_{2}}\And {{B}_{2}}=(72-36)\mu C\] \[{{Q}_{2}}=36\mu C\] \[\therefore \]\[{{A}_{1}}\] and \[{{B}_{1}}\]have \[-\upsilon e\] charge and \[{{A}_{2}}{{B}_{2}}\]have \[+\upsilon e.\] So combination is in parallel \[Ceq.=9\mu F\] V across each\[V=\frac{{{Q}_{1}}\times {{Q}_{2}}}{C}=\frac{36\mu C}{9\mu F}=4\]volts