
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) \[\dfrac{C}{2},2V\]
Answer
123.9k+ views
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
Note:
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.
\[Q = CV\]
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
Note:
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.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

Difference Between Circuit Switching and Packet Switching

Difference Between Mass and Weight

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips

Sign up for JEE Main 2025 Live Classes - Vedantu

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main Login 2045: Step-by-Step Instructions and Details

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Physics Average Value and RMS Value JEE Main 2025

JEE Mains 2025 Correction Window Date (Out) – Check Procedure and Fees Here!
