Electrostatics Question 142
Question: A 40 mF capacitor in a defibrillator is charged to 3000 V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2ms. The power delivered to the patient is [AIIMS 2004]
Options:
A) 45 kW
B) 90 kW
C) 180 kW
D) 360 kW
Show Answer
Answer:
Correct Answer: B
Solution:
Power $ =\frac{\frac{1}{2}CV^{2}}{t}=\frac{1\times 40\times {{10}^{-6}}\times {{(3000)}^{2}}}{2\times 2\times {{10}^{-3}}}=90kW $
