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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Physics
Chapter
Capacitance

BITSAT 2011 · Physics · Capacitance

If the potential of a capacitor having capacity 6 μ F is increased...

If the potential of a capacitor having capacity 6μF6 \mu \mathrm{F} is increased from 10V10 \mathrm{ V} to 20V20 \mathrm{ V}, then increase in its energywill be

Options

  1. A4×10−4J4 \times 10^{-4} \mathrm{ J}
  2. B4×10−6J4 \times 10^{-6} \mathrm{ J}
  3. C9×10−4J9 \times 10^{-4} \mathrm{ J}
  4. D12×10−6J12 \times 10^{-6} \mathrm{ J}

Correct answer

C. 9×10−4J9 \times 10^{-4} \mathrm{ J}

Step-by-step solution

Capacitance of capacitor (C) =6μF=6×10−6F=6 \mu \mathrm{F}=6 \times 10^{-6} \mathrm{ F}; Initial potential \left(\mathrm{V}_{1}\right)=10 \mathrm{ V}andfinalpotentialand final potential\left(\mathrm{V}_{2}\right)=20 \mathrm{ V}.Theincreaseinenergy. The increase in energy(\Delta \mathrm{U})=12C(V22−V12)=12×(6×10−6)×[(20)2−(10)2]=(3×10−6)×300=9×10−4J.\begin{array}{l} =\frac{1}{2} \mathrm{C} \left(\mathrm{V}_{2}^{2}-\mathrm{V}_{1}^{2}\right)\\ =\frac{1}{2} \times \left(6 \times10^{-6}\right)\times \left[(20)^{2}-(10)^{2}\right] \\ =\left(3 \times10^{-6}\right)\times300=9 \times10^{-4} \mathrm{ J} . \end{array}
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