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JEE Main
Exam
JEE Main
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Physics
Chapter
Capacitance

JEE Main · Physics · Capacitance

A body of capacity 4 μ F is charged to 80 ~V and another body of...

A body of capacity 4μF4 \mu \mathrm{F} is charged to 80 V80 \mathrm{~V} and another body of capacity 6μF6 \mu \mathrm{F} is charged to 30 V30 \mathrm{~V}. When they are connected the energy lost by 4μF4 \mu \mathrm{F} capacitor is

Options

  1. A9.8 mJ9.8 \mathrm{~mJ}
  2. B4.6 mJ4.6 \mathrm{~mJ}
  3. C3.2 mJ3.2 \mathrm{~mJ}
  4. D2.5 mJ2.5 \mathrm{~mJ}

Correct answer

A. 9.8 mJ9.8 \mathrm{~mJ}

Step-by-step solution

C2=4μF,V1=80\mathrm{C}_2=4 \mu \mathrm{F}, V_1=80 volt, C2=6μF,V2=30C_2=6 \mu \mathrm{F}, V_2=30 volt Energy loss, ΔU=12C1C2C2+C2(V1−V2)2=12×4×10−6×6×10−44×10−6+6×10−4(80−30)2=12×10−610×2500=1.2×25×10−4=30×10−4=3×10−3 J\begin{aligned} \Delta U & =\frac{1}{2} \frac{C_1 C_2}{C_2+C_2}\left(V_1-V_2\right)^2 \\ & =\frac{1}{2} \times \frac{4 \times 10^{-6} \times 6 \times 10^{-4}}{4 \times 10^{-6}+6 \times 10^{-4}}(80-30)^2 \\ & =\frac{12 \times 10^{-6}}{10} \times 2500=1.2 \times 25 \times 10^{-4} \\ & =30 \times 10^{-4}=3 \times 10^{-3} \mathrm{~J}\end{aligned} Initial energy of 4μF4 \mu \mathrm{F} capacitor. En=12C1V12=12×4×10−4×(80)2=2×10−4×6400=12.8×10−3 J\begin{aligned} E_{\mathrm{n}} & =\frac{1}{2} C_1 V_1^2 \\ & =\frac{1}{2} \times 4 \times 10^{-4} \times(80)^2 \\ & =2 \times 10^{-4} \times 6400 \\ & =12.8 \times 10^{-3} \mathrm{~J}\end{aligned} Energy lost by C1=12.8×10−1−3×10−1C_1=12.8 \times 10^{-1}-3 \times 10^{-1}=9.8×10−3 J=9.8 mJ\begin{aligned} & =9.8 \times 10^{-3} \mathrm{~J} \\ & =9.8 \mathrm{~mJ}\end{aligned}$$
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