A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 50.0 - \(\mu \mathrm{F}\) capacitor is charged to \(6.0 \mathrm{kV}\). Paddles are used to make an electric connection to the patient's chest. A pulse of current lasting 1.0 ms partially discharges the capacitor through the patient. The electrical resistance of the patient (from paddle to paddle) is \(240 \Omega\) (a) What is the initial energy stored in the capacitor? (b) What is the initial current through the patient? (c) How much energy is dissipated in the patient during the 1.0 ms? (d) If it takes 2.0 s to recharge the capacitor, compare the average power supplied by the power source with the average power delivered to the patient. (e) Referring to your answer to part (d), explain one reason a capacitor is used in a defibrillator.

Step by step solution

01

Calculate Capacitor Energy

To find the energy stored in the capacitor, we will use the formula: \(E_c = \frac{1}{2}CV^2\). Plug in the values of the capacitance, \(C = 50.0 \times 10^{-6} \, F\), and voltage, \(V = 6.0 \times 10^3 \, V\), into the formula: \(E_c = \frac{1}{2}(50.0 \times 10^{-6}\,\text{F})(6.0 \times 10^3\,\text{V})^2\)

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