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Two capacitors each having a capacitor $C$ and breakdown voltage $V$ are joined in series. The effective capacitance and maximum working voltage of the combination is:
(A) $2C,2V$
(B) $\dfrac{C}{2},\dfrac{V}{2}$
(C) $2C,V$
(D) \[\dfrac{C}{2},2V\]

Last updated date: 13th Jun 2024
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Hint: We know for a given capacitor, charge \[Q\] on a capacitor is proportional to potential difference \[V\], between the plates.
\[Q = CV\]
And, for series combination charges on capacitors remain the same.

Complete Step by Step Answer
Figure, shows two capacitors connected in series. The capacitance is \[C\] and \[C\].

Now, let us take the potential of the right plate of the second plate to be zero. The potential of the left plate of the first capacitor is \[E\]. Since, the breakdown voltage of capacitors is \[V\]. Therefore capacitor \[1\],
\[E - V = \dfrac{Q}{C}...........(i)\]
Similarly, for other capacitor,
\[V - 0 = \dfrac{Q}{C}...........(ii)\]
Adding equation \[(i)\] and \[(ii)\]
\[E = Q\left( {\dfrac{1}{C} + \dfrac{1}{C}} \right)\,.........\,(iii)\]
If the equivalent capacitance of the combination is \[{C_{eq}}\].
\[{C_{eq}} = \dfrac{Q}{E}\,..........\,(iv)\]
Using equation \[(iii)\] and \[(iv)\] we get,
\[{C_{eq}} = \dfrac{C}{2}\,\]
And, the maximum working voltage is \[E\].
Hence, \[E = V + V = 2V\]

Hence, Option (D) is correct

Charge on series combination remains same but voltage changes with respect to the capacitance whereas voltage on parallel combination remains same but charge varies in accordance to capacitance.