Question 16.10: TYPICAL VOLTAGE, ENERGY, AND DISCHARGE TIME FOR A DEFIBRILLA...
TYPICAL VOLTAGE, ENERGY, AND DISCHARGE TIME FOR A DEFIBRILLATOR
GOAL Apply energy and power concepts to a capacitor.
PROBLE M A fully charged defibrillator contains 1.20 \mathrm{~kJ} of energy stored in a 1.10 \times 10^{-4} \mathrm{~F} capacitor. In a discharge through a patient, 6.00 \times 10^{2} \mathrm{~J} of electrical energy is delivered in 2.50 \mathrm{~ms}. (a) Find the voltage needed to store 1.20 \mathrm{~kJ} in the unit. (b) What average power is delivered to the patient?
STRATEGY Because we know the energy stored and the capacitance, we can use Equation 16.17
P E_{C} =\frac{1}{2} Q \Delta V =\frac{1}{2} C(\Delta V)^2 = \frac{Q^2}{2C} [16.17]
to find the required voltage in part (a). For part (b), dividing the energy delivered by the time gives the average power.
(a) Find the voltage needed to store 1.20 \mathrm{~kJ} in the unit.
Solve Equation 16.17 for ΔV:
\begin{aligned}P E_{C} &=\frac{1}{2} C \Delta V^{2} \\\Delta V &=\sqrt{\frac{2 \times P E_{C}}{C}} \\&=\sqrt{\frac{2\left(1.20 \times 10^{3} \mathrm{~J}\right)}{1.10 \times 10{ }^{-4} \mathrm{~F}}} \\&=4.67 \times 10^{3} \mathrm{~V}\end{aligned}
(b) What average power is delivered to the patient?
Divide the energy delivered by the time:
\begin{aligned}P_{\mathrm{av}} &=\frac{\text { energy delivered }}{\Delta t}=\frac{6.00 \times 10^{2} \mathrm{~J}}{2.50 \times 10^{-3} \mathrm{~s}} \\&=2.40 \times 10^{5} \mathrm{~W}\end{aligned}
REMARKS The power delivered by a draining capacitor isn’t constant, as we’ll find in the study of R C circuits in Topic 18 . For that reason, we were able to find only an average power. Capacitors are necessary in defibrillators because they can deliver energy far more quickly than batteries. Batteries provide current through relatively slow chemical reactions, whereas capacitors can quickly release charge that has already been produced and stored.