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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Physics
Chapter
Capacitance

BITSAT 2013 · Physics · Capacitance

Two identical capacitors, have the same capacitance C . One of them...

Two identical capacitors, have the same capacitance C\mathrm{C}. One of them is charged to potential V1\mathrm{V}_{1} and the other to V2\mathrm{V}_{2}. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is -

Options

  1. A14C(V12−V22)\frac{1}{4} \mathrm{C} \left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)
  2. B14C(V12+V22)\frac{1}{4} C \left(V_{1}^{2}+V_{2}^{2}\right)
  3. C\frac{1}{4} \mathrm{C}(V1−V2)2\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}
  4. D\frac{1}{4} C(V1+V2)2\left(V_{1}+V_{2}\right)^{2}

Correct answer

C. \frac{1}{4} \mathrm{C}(V1−V2)2\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}

Step-by-step solution

Initial energy of combined system U1=12CV12+12CV22\mathrm{U}_{1}=\frac{1}{2} \mathrm{CV}_{1}^{2}+\frac{1}{2} \mathrm{CV}_{2}^{2} Final common potential, V=V1+V22,\mathrm{V}=\frac{\mathrm{V}_{1}+\mathrm{V}_{2}}{2}, Final energy of system, U2=2×12C(V1+V22)2\mathrm{U}_{2}=2 \times \frac{1}{2} \mathrm{C}\left(\frac{\mathrm{V}_{1}+\mathrm{V}_{2}}{2}\right)^{2} Hence loss of energy =U1−U2=14C(V1−V2)2=\mathrm{U}_{1}-\mathrm{U}_{2}=\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}
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