Energy and power of a defibrillator

Problem

A defibrillator uses a capacitor of $C = 50\;\mu\text{F}$ charged to $V = 1{,}6\;\text{kV}$. (a) Determine the stored energy $U$. (b) If the discharge onto the chest lasts $\Delta t \approx 4\;\text{ms}$, estimate the average power delivered in kW. (c) Comment on why, even though the energy is modest compared to a meal, the effects on the heart are so pronounced.

Solution

Data in SI units. $C = 50\;\mu\text{F} = 5{,}0\cdot 10^{-5}\;\text{F}$, $V = 1{,}6\;\text{kV} = 1{,}6\cdot 10^{3}\;\text{V}$, $\Delta t = 4\;\text{ms} = 4\cdot 10^{-3}\;\text{s}$.

(a) Stored energy. $U = \tfrac12\,C V^2 = \tfrac12\,(5{,}0\cdot 10^{-5})\,(1{,}6\cdot 10^{3})^2 = \tfrac12\,(5{,}0\cdot 10^{-5})\,(2{,}56\cdot 10^{6})$ $\ev{U \approx 64\;\text{J}}$

(b) Average power. Power is the energy delivered over the discharge time: $P_{\text{media}} = \frac{U}{\Delta t} = \frac{64\;\text{J}}{4\cdot 10^{-3}\;\text{s}} = 1{,}6\cdot 10^{4}\;\text{W}$ $\ev{P_{\text{media}} \approx 16\;\text{kW}}$

(c) Comment. An energy of 64 J is negligible on a dietary scale (a bite of apple contains thousands of times more). What makes the effect dramatic is the power: those 64 J are released in just 4 milliseconds, generating an instantaneous power of tens of kilowatts. It is the speed of the discharge — not the total amount of energy — that produces the current pulse capable of resetting the heart’s electrical rhythm.

Links

Topics: Electric field and potential Concepts: Capacitance and capacitor · Electrostatic potential energy Objects: Parallel-plate capacitor