The internal circuitry of a heart defibrillator includes a 7.42 µF parallel plate capacitor. When Hawkeye Pierce applies the defibrillator's paddles across a patient's torso, the result is a series RC circuit. If the capacitor's voltage declines from 10.7 kV to 213 V in 0.049 s, what must be the resistance (in 2) of the circuit through the patient's torso?

The internal circuitry of a heart defibrillator includes a 7.42 µF parallel plate capacitor. When Hawkeye Pierce applies the defibrillator's paddles across a patient's torso, the result is a series RC circuit. If the capacitor's voltage declines from 10.7 kV to 213 V in 0.049 s, what must be the resistance (in 2) of the circuit through the patient's torso?

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter29: Direct Current (dc) Circuits

Section: Chapter Questions

Problem 19PQ

See similar textbooks

Similar questions

The values of the components in a simple series RC circuit containing a switch (Fig. P21.53) are C = 1.00 F, R = 2.00 106 , and = 10.0 V. At the instant 10.0 s after the switch is closed, calculate (a) the charge on the capacitor, (b) the current in the resistor, (c) the rate at which energy is being stored in the capacitor, and (d) the rate at which energy is being delivered by the battery.
Figure P29.60 shows a simple RC circuit with a 2.50-F capacitor, a 3.50-M resistor, a 9.00-V emf, and a switch. What are a. the charge on the capacitor, b. the current in the resistor, c. the rate at which the capacitor is storing energy, and d. the rate at which the battery is delivering energy exactly 7.50 s alter the switch is closed?
Consider a series RC circuit as in Figure P18.35 for which R = 1.00 M, C = 5.00 F, and = 30.0 V. Find (a) the time constant of the circuit and (b) the maximum charge on the capacitor after the switch is thrown closed. (c) Find the current in the resistor 10.0 s after the switch is closed. Figure P18.35 Problem 35 and 38.
The label on a battery-powered radio recommends the use of a rechargeable nickel-cadmium cell (nicads), although it has a 1.25-V emf, whereas an alkaline cell has a 1.58-V emf. The radio has a 3.20 resistance.(a) Draw a circuit diagram of the radio and its battery. Now, calculate the power delivered to the radio (b) when using a nicad cells, each having an internal resistance of 0.0400 , and (c) when using an alkaline cell, having an internal resistance of 0.200 . (d) Does this differenceseem significant, considering that the radio's effective resistance is lowered when its volume is aimed up?
Consider a series RC circuit as in Figure P28.38 for which R = 1.00 M, C = 5.00 F, and = 30.0 V. Find (a) the time constant of the circuit and (b) the maximum charge on the capacitor after the switch is thrown closed. (c) Find the current in the resistor 10.0 s after the switch is closed.
An automobile starter motor has an equivalent resistance of 0.0500 and is supplied by a 12.0-V battery with a 0.0100- internal resistance, (a) What is thecurrent to the motor? (b) What voltage is applied to it? (c) What power is supplied to the motor? (d) Repeat these calculations for when the battery connections are corroded and add 0.0900 to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)
In the RC circuit shown in Figure P29.78, an ideal battery with emf and internal resistance r is connected to capacitor C. The switch S is initially open and the capacitor is uncharged. At t = 0, the switch is closed. a. Determine the charge q on the capacitor at time t. b. Find the current in the branch be at time t. What is the current as t goes to infinity?
Figure P18.26 shows a voltage divider, a circuit used to obtain a desired voltage Vout from a source voltage . Determine the required value of R2 if = 5.00 V, Vout = 1.50 V and R1 = 1.00 103 (Hint: Use Kirchhoff's loop rule, substituting Vout = IR2, to find the current. Then solve Ohms law for R2. Figure P18.26
A battery with = 6.00 V and no internal resistance supplies current to the circuit shown in Figure P27.9. When the double-throw switch S is open as shown in the figure, the current in the battery is 1.00 mA. When the switch is closed in position a, the current in the battery is 1.20 mA. When the switch is closed in position b, the current in the battery is 2.00 mA. Find the resistances (a) R1, (b) R2, and (c) R3. Figure P27.9 Problems 9 and 10.
Three identical 60.0-W, 120-V lightbulbs are connected across a 120-V power source as shown in Figure P28.72. Assuming the resistance of each lightbulb is constant (even though in reality the resistance might increase markedly with current), find (a) the total power supplied by the power source and (b) the potential difference across each lightbulb.
For the circuit shown in Figure P28.55. the ideal voltmeter reads 6.00 V and the ideal ammeter reads 3.00-k. Find (a) the value of K, (b) the emf of the battery, and (c) the voltage across the 3.00-kft resistor.
Figure P18.26 shows a voltage divider, a circuit used to obtain a desired voltage Vout from a source voltage . Determine the required value of R2 if = 5.00 V, Vout = 1.50 V and R1 = 1.00 103 (Hint: Use Kirchhoff's loop rule, substituting Vout = IR2, to find the current. Then solve Ohms law for R2. Figure P18.26