Question

A $40\mu F$ capacitor in a defibrillator is charged to $3000V$. The energy stored in the capacitor is sent through the patient during a pulse of duration $2$ ms. The power delivered to the patient is :

• A. $30W$
• B. $90kW$
• C. $180kW$
• D. $360kW$

Answer 1

Answer: (A)

$30W$

Given, $c=40\mu F=40\times {10}^{-6}FV=3000V$

Energy sent through the patient during a pulse of duration $=2ms$

The power delivered $P=VI=\frac{E}{t}=\frac{\frac{1}{2}c{v}^2}{t}=\frac{\frac{1}{2}\times 40\times {10}^{-6}\times 3000}{2\times {10}^{-3}}=\frac{1}{2}\times \frac{1}{2}\times 40\times {10}^{-3}\times 3000=\frac{1}{4}\times 40\times {10}^{-3}\times 3\times {10}^3=30W$