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Capacitor question

2023 · 30 Jan · Shift 1 · Q68
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  5. /2023 · 30 Jan · Shift 1 · Q68

Capacitor question

2023 · 30 Jan · Shift 1 · Q68

JEE MainPhysicsCapacitorNumerical+4 / −1
A capacitor of capacitance 900μF900 \mu \mathrm{F}900μF is charged by a 100 V100 \mathrm{~V}100 V battery. The capacitor is disconnected from the battery and connected to another uncharged identical capacitor such that one plate of uncharged capacitor connected to positive plate and another plate of uncharged capacitor connected to negative plate of the charged capacitor. The loss of energy in this process is measured as x×10−2 Jx \times 10^{-} { }^{2} \mathrm{~J}x×10−2 J. The value of xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 225

  1. Initial charge and energy on the charged capacitor

Given:

  • Capacitance of each capacitor: C=900 μF=900×10−6 FC = 900\,\mu\text{F} = 900 \times 10^{-6}\,\text{F}C=900μF=900×10−6F
  • Initial voltage on first capacitor: V=100 VV = 100\,\text{V}V=100V

Initial energy stored in the charged capacitor is

Ui=12CV2U_i = \frac{1}{2}CV^2Ui​=21​CV2

Substitute values:

Ui=12(900×10−6)(100)2U_i = \frac{1}{2}(900 \times 10^{-6})(100)^2Ui​=21​(900×10−6)(100)2 Ui=12(900×10−6)(10000)U_i = \frac{1}{2}(900 \times 10^{-6})(10000)Ui​=21​(900×10−6)(10000) Ui=12(9)=4.5 JU_i = \frac{1}{2}(9) = 4.5\,\text{J}Ui​=21​(9)=4.5J
  1. After connecting to an identical uncharged capacitor

The two capacitors are identical and connected with like plates together:

  • positive to positive
  • negative to negative

So charge redistributes equally.

Initial charge on the charged capacitor:

Q=CV=(900×10−6)(100)=0.09 CQ = CV = (900 \times 10^{-6})(100) = 0.09\,\text{C}Q=CV=(900×10−6)(100)=0.09C

Since the second capacitor is identical and initially uncharged, final voltage on both capacitors becomes:

Vf=Q2C=CV2C=V2=50 VV_f = \frac{Q}{2C} = \frac{CV}{2C} = \frac{V}{2} = 50\,\text{V}Vf​=2CQ​=2CCV​=2V​=50V
  1. Final energy of the system

Now there are two capacitors, each with capacitance CCC and voltage 50 V50\,\text{V}50V.

Energy on each capacitor:

Ueach=12CVf2U_{each} = \frac{1}{2}C V_f^2Ueach​=21​CVf2​

Total final energy:

Uf=2⋅12CVf2=CVf2U_f = 2 \cdot \frac{1}{2}C V_f^2 = C V_f^2Uf​=2⋅21​CVf2​=CVf2​ Uf=(900×10−6)(50)2U_f = (900 \times 10^{-6})(50)^2Uf​=(900×10−6)(50)2 Uf=(900×10−6)(2500)U_f = (900 \times 10^{-6})(2500)Uf​=(900×10−6)(2500) Uf=2.25 JU_f = 2.25\,\text{J}Uf​=2.25J
  1. Loss of energy
ΔU=Ui−Uf=4.5−2.25=2.25 J\Delta U = U_i - U_f = 4.5 - 2.25 = 2.25\,\text{J}ΔU=Ui​−Uf​=4.5−2.25=2.25J

Given that loss of energy is written as:

x×10−2 Jx \times 10^{-2}\,\text{J}x×10−2J

So,

x×10−2=2.25x \times 10^{-2} = 2.25x×10−2=2.25 x=2.2510−2=225x = \frac{2.25}{10^{-2}} = 225x=10−22.25​=225
  1. Final answer
x=225\boxed{x = 225}x=225​

The derived answer matches the stored correct answer.

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