Figure P18.37 shows a simplified model of a cardiac defibrillator, a device used to patients in ventricular fibrillation. When the switch S is toggled to the left, the capacitor C charges through the resistor R .When the switch is toggled to the right, the capacitor discharges current through the patient’s torso, modeled as the resistor R torso , allowing the heart’s normal rhythm to be reestablished. (a) If the capacitor is initially uncharged with C = 8.00 µ F and ε = 1250 V, find the value of R required to charge the capacitor to a voltage of 775 V in 1.50 s. (b) If the capacitor is then discharged across the patient’s torso with, R torso = 1250 Ω, calculate the voltage across the capacitor after 5.00 ms. Figure P18.37
Figure P18.37 shows a simplified model of a cardiac defibrillator, a device used to patients in ventricular fibrillation. When the switch S is toggled to the left, the capacitor C charges through the resistor R .When the switch is toggled to the right, the capacitor discharges current through the patient’s torso, modeled as the resistor R torso , allowing the heart’s normal rhythm to be reestablished. (a) If the capacitor is initially uncharged with C = 8.00 µ F and ε = 1250 V, find the value of R required to charge the capacitor to a voltage of 775 V in 1.50 s. (b) If the capacitor is then discharged across the patient’s torso with, R torso = 1250 Ω, calculate the voltage across the capacitor after 5.00 ms. Figure P18.37
Solution Summary: The author explains the formula to calculate time constant tau and the value of resistor R to charge the capacitor.
Figure P18.37 shows a simplified model of a cardiac defibrillator, a device used to patients in ventricular fibrillation. When the switch S is toggled to the left, the capacitor C charges through the resistor R .When the switch is toggled to the right, the capacitor discharges current through the patient’s torso, modeled as the resistor Rtorso, allowing the heart’s normal rhythm to be reestablished. (a) If the capacitor is initially uncharged with C = 8.00 µF and ε = 1250 V, find the value of R required to charge the capacitor to a voltage of 775 V in 1.50 s. (b) If the capacitor is then discharged across the patient’s torso with, Rtorso = 1250 Ω, calculate the voltage across the capacitor after 5.00 ms.
A capacitor with a capacitance of C = 5.95×10−5 F is charged by connecting it to a 12.5 −V battery. The capacitor is then disconnected from the battery and connected across an inductor with an inductance of L = 1.55 H . At the time 2.35×10−2 s after the connection to the inductor is made, what is the current in the inductor? At that time, how much electrical energy is stored in the inductor?
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What is Electromagnetic Induction? | Faraday's Laws and Lenz Law | iKen | iKen Edu | iKen App; Author: Iken Edu;https://www.youtube.com/watch?v=3HyORmBip-w;License: Standard YouTube License, CC-BY