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AIIMS 2004
Exam
AIIMS
Year
2004
Subject
Physics
Chapter
Capacitance

AIIMS 2004 · Physics · Capacitance

A 40 μ F capacitor in a defibrillator is charged to 3000 ~V . The...

A 40μF40 \mu \mathrm{F} capacitor in a defibrillator is charged to 3000 V3000 \mathrm{~V}. The energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms2 \mathrm{~ms}. The power delivered to the patient is

Options

  1. A45 kW45 \mathrm{~kW}
  2. B90 kW90 \mathrm{~kW}
  3. C180 kW180 \mathrm{~kW}
  4. D360 kW360 \mathrm{~kW}.

Correct answer

B. 90 kW90 \mathrm{~kW}

Step-by-step solution

The energy stored in capacitor =12\frac{1}{2} C V^2 =12×40×10−6×(3000)2=20×9×106×10−6=180 J.\begin{aligned}& =\frac{1}{2} \times40 \times10^{-6} \times(3000)^2 \\ & =20 \times9 \times10^6 \times10^{-6}=180 \mathrm{~J}. \end{aligned}The power deliveredin2in 2\mathrm{~ms}is=is =\frac{180}{2 \times 10^{-3}}=9090\times10310^3\mathrm{~W}=9090\mathrm{~kW}$ .
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