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CBSE Class 12 Physics 2017 Delhi Set 3 Paper

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Question : 7 of 8
Marks: +1, -0
Two identical capacitors of 12pF12 \mathrm{pF} each are connected in series across a battery of 50V50 \mathrm{V}. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy will be stored in the combination now?
Also find the charge drawn from the battery in each case.
Solution:  
Equivalent capacitance in series
Energy in series combination
Charge in series combination
Equivalent capacitance in parallel combination
Energy in parallel combination
Charge in parallel combination
In series combination :
  1Cs  =(  112+  112)(pF)−1\;\frac{1}{C_s}\;= \left(\;\frac{1}{12}+\;\frac{1}{12}\right)(\mathrm{pF})^{-1}
Cs  =6×10−12pFC_s\;=6 \times 10^{-12} \mathrm{pF}
Us  =  12CsV2U_s\;=\;\frac{1}{2} C_s V^2
Us  =  12×6×10−12×50×50JU_s\;=\;\frac{1}{2} \times 6 \times 10^{-12} \times 50 \times 50 \mathrm{J}
Us  =75×10−10JU_s\;=75 \times 10^{-10} \mathrm{J}
qs  =CsVq_s\;=C_s V
  =6×10−12×50\;=6 \times 10^{-12} \times 50
  =300×10−12C=3×10−10C\;=300 \times 10^{-12} \mathrm{C}=3 \times 10^{-10} \mathrm{C}
In parallel combination :
  Cp=(12+12)pF\;C_p=(12+12) \mathrm{pF}
∴    Cp=24×10−12F\therefore\;\;C_p=24 \times 10^{-12} \mathrm{F}
  Up=  12×24×10−12×2500J\;U_p=\;\frac{1}{2} \times 24 \times 10^{-12} \times 2500 \mathrm{J}
  Us=3×10−8J\;U_s=3 \times 10^{-8} \mathrm{J}
  qp=CpV\;q_p=C_p V
  qp=24×10−12×50C\;q_p=24 \times 10^{-12} \times 50 \mathrm{C}
  qp=1.2×10−9C\;q_p=1.2 \times 10^{-9} \mathrm{C}
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