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QJEEM15Ph023c: JEE Main 2015 Physics - Class 12 JEEM15Ph023c Solution: In the given circuit, charges on the capacitor changes as is…

In the given circuit, charges Q2{Q_2} on the 2μF2\mu F capacitor changes as CC is varied from 1 μF1\,\mu F to 3μF.3\mu F. Q2{Q_2} as a function of ′C′'C' is given properly by:
(figures  are  drawn  schematically  and  are  not  to  scale)\left( {figures\,\,are\,\,drawn\,\,schematically\,\,and\,\,are\,\,not\,\,to\,\,scale} \right)

JEE Main 2015 (Offline) Physics - Capacitor Question 140 English

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/13af33ec9a4f3abea9a5699b4d84ed1be185abac9f8bc2c046f0c3a042016f56.webp
  • B https://ik.imagekit.io/novatra/photos/98927542c97cc7f560a649189f034071a7efe6c77bb0f0f0d74fff88300a8ade.webp
  • C https://ik.imagekit.io/novatra/photos/9b5e289c4739640260572ce9a9e7166c0f0b412a30f6ba0e7d3db80e4a322142.webp
  • D https://ik.imagekit.io/novatra/photos/9b114747e7aa628f856cf862e904da4583404940f5a61351fed470082d17219b.webp

Step-by-Step Verified Solution

JEE Main 2015 (Offline) Physics - Capacitor Question 140 English Explanation 1

From figure, Q2=22+1Q=23Q{Q_2} = {2 \over {2 + 1}}Q = {2 \over 3}Q

Q=E(C×3C+3)Q = E\left( {{{C \times 3} \over {C + 3}}} \right)

∴\therefore Q2=23(3CEC+3)=2CEC+3{Q_2} = {2 \over 3}\left( {{{3CE} \over {C + 3}}} \right) = {{2CE} \over {C + 3}}

Therefore graph dd correctly dipicts.

JEE Main 2015 (Offline) Physics - Capacitor Question 140 English Explanation 2

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