A defibrillator is designed to pass a large current through a patient’s torso in order to stop dangerous heart rhythms. Its key part is a capacitor that is charged to a high voltage. The patient’s torso plays the role of a resistor in an RC circuit. When a switch is closed, the capacitor discharges through the patient’s torso. A jolt from a defibrillator is intended to be intense and rapid; the maximum current is very large, so the capacitor discharges quickly. This rapid pulse depolarizes the heart, stopping all electrical activity. This allows the heart’s internal nerve circuitry to reestablish a healthy rhythm. A typical defibrillator has a 32 μ F capacitor charged to 5000 V. The electrodes connected to the patient are coated with a conducting gel that reduces the resistance of the skin to where the effective resistance of the patient’s torso is 100 Ω. In some cases, the defibrillator may be charged to a lower voltage. How will this affect the time constant of the discharge? A. The time constant will increase. B. The time constant will not change. C. The time constant will decrease.
A defibrillator is designed to pass a large current through a patient’s torso in order to stop dangerous heart rhythms. Its key part is a capacitor that is charged to a high voltage. The patient’s torso plays the role of a resistor in an RC circuit. When a switch is closed, the capacitor discharges through the patient’s torso. A jolt from a defibrillator is intended to be intense and rapid; the maximum current is very large, so the capacitor discharges quickly. This rapid pulse depolarizes the heart, stopping all electrical activity. This allows the heart’s internal nerve circuitry to reestablish a healthy rhythm. A typical defibrillator has a 32 μ F capacitor charged to 5000 V. The electrodes connected to the patient are coated with a conducting gel that reduces the resistance of the skin to where the effective resistance of the patient’s torso is 100 Ω. In some cases, the defibrillator may be charged to a lower voltage. How will this affect the time constant of the discharge? A. The time constant will increase. B. The time constant will not change. C. The time constant will decrease.
A defibrillator is designed to pass a large current through a patient’s torso in order to stop dangerous heart rhythms. Its key part is a capacitor that is charged to a high voltage. The patient’s torso plays the role of a resistor in an RC circuit. When a switch is closed, the capacitor discharges through the patient’s torso. A jolt from a defibrillator is intended to be intense and rapid; the maximum current is very large, so the capacitor discharges quickly. This rapid pulse depolarizes the heart, stopping all electrical activity. This allows the heart’s internal nerve circuitry to reestablish a healthy rhythm.
A typical defibrillator has a 32 μF capacitor charged to 5000 V. The electrodes connected to the patient are coated with a conducting gel that reduces the resistance of the skin to where the effective resistance of the patient’s torso is 100 Ω.
In some cases, the defibrillator may be charged to a lower voltage. How will this affect the time constant of the discharge?
RC Circuits: For the circuit shown in the figure, C = 13 µF and R = 7.6 MQ. Initially the switch S is
open with the capacitor charged to a voltage of 80 V. The switch is then closed at time t = 0.00 s.
What is the charge on the capacitor 40 s after closing the switch?
3100 μC
2500 μC
2700 μC
3300 μC
2900 μC
R
Chapter 23 Solutions
College Physics: A Strategic Approach (3rd Edition)
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