A defibrillator has a 130-$\mu$F capacitor charged to 2000 V and an internal resistance of 30$\Omega$. If the resistance of the path of the discharge through the patient is 170$\Omega$, how much energy is delivered to the patient upon complete discharge of the capacitor?
A) 170 J
B) 30 J
C) 220 J
D) 240 J

Question

Quizplus · Accepted Answer

The energy stored in the capacitor is given by the formula $ E = \frac{1}{2} C V^2 $. With $ C = 130 \mu F = 130 \times 10^{-6} F $ and $ V = 2000 V $, the total energy stored is $ E = \frac{1}{2} \times 130 \times 10^{-6} \times (2000)^2 = 260 J $. The total resistance in the circuit is $ 30 \Omega + 170 \Omega = 200 \Omega $. The fraction of energy delivered to the patient is given by the ratio of the patient's resistance to the total resistance, $ \frac{170}{200} $. Therefore, the energy delivered is $ 260 \times \frac{170}{200} = 221 J $, which is closest to 220 J.

A Defibrillator Has a 130- μ\muμ F Capacitor Charged to 2000 V and an Internal Resistance

A Defibrillator Has a 130- $\mu$ F Capacitor Charged to 2000 V and an Internal Resistance