Answer to Question #121650 in Mechanics | Relativity for nicolee heath

Question

A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 50.0x10-6 F capacitor is charged to 6.00x103 V. Paddles are used to make an electrical connection to the patient's chest. The electrical resistance of the patient (from paddle to paddle) is 245 Ω. A pulse of current lasting 1.50x10-3 s partially discharges the capacitor, i.e., the switch is closed for 1.50x10-3 s and then opened.

What is the time constant of this circuit?
What is the energy initially stored in the capacitor?
What is the maximum charge on the capacitor before it is discharged?
How much energy is dissipated in the patient (resistor) during the 1.50x10-3 s that the switch is closed?

Answer 1

$С=50\times10^{-6}F\\U=6\times10^3V\\R=245\Omega\\t=50\times10^{-3}s;\\R\times C-?\\W-?\\Q-?\\energy\; is \;dissipated\\W(total)-W(t)(remaining)-?;\\Solution:\\1)\;R\times C=245\times50\times 10^{-6}=0.01225s\\2)\;W=\frac{C\times U^2}{2}=\frac{50\times10^{-6}\times 36\times10^6}{2}=900J\\3)\;Q=C\times U=50\times 10^{-6}\times6\times10^3=0.3C\\4)W-W(t)=W-\frac{{C\times( U\times e^{-\frac{t}{R\times C}}})^2}{2}=\\=900- 900\times e^{-8.16} =\\=900-0.261=899.74J$