Question

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

• A. $45\text{kW}$
• B. $90\text{kW}$
• C. $180\text{kW}$
• D. $360\text{kW}$

Answer 1

The correct option is B $90\text{kW}$

Energy delivered by the defibrillator is equal to enery stored in the capacitor and its value is given by

$U=\frac{1}{2}C{V}^2=\frac{1}{2}\times 40\times {10}^{-6}\times {3000}^2=180\text{J}$

Pulse duration = Time taken for energy to be delivered = $2\text{ms}=2\times {10}^{-3}\text{s}$

$\therefore$Power delivered will be equal to

$P=\frac{\text{Energy delivered}}{\text{Time taken}}=\frac{180}{2\times {10}^{-3}}$

$\Rightarrow P=90,000\text{W}=90\text{kW}$

Hence, option $\left(b\right)$ is correct.

Key concept - Power is the rate at which work is done.