A defibrillator passes a brief burst of current through the...

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 $1.0 \mathrm{ms}$ 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 $1.0 \mathrm{ms} ?$ (d) If it takes $2.0 \mathrm{s}$ 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.

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 $1.0 \mathrm{ms}$ 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 $1.0 \mathrm{ms} ?$
(d) If it takes $2.0 \mathrm{s}$ 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.

Key Concepts

Average Power and Duty Cycle

Average power in a circuit is defined as the total energy delivered over a period of time divided by that period. In systems with short, high-power pulses followed by long recharge intervals, the average power is significantly lower than the instantaneous power delivered during the pulse. This concept highlights the importance of timing and energy management in power supply design.

Role of Capacitors in High-Power Applications

Capacitors are often used in high-power applications like defibrillators because they can rapidly store and then release a large amount of energy in a very short time. This capability allows a low-power supply to build up energy over time and discharge it quickly, ensuring that the necessary high peak power is available when needed. Additionally, capacitors help isolate the charging circuit from the high-energy discharge, thereby improving safety and controllability.

Energy Dissipation in a Resistor

When current flows through a resistor, electrical energy is converted into heat, a process known as energy dissipation. In pulse circuits, calculating the energy dissipated in the resistor involves integrating the power over the duration of the pulse. This concept is important for understanding how much energy is effectively transferred to or lost in a load during discharge.

Capacitor Energy Storage

A capacitor is an electrical component that stores energy in the form of an electric field. The energy stored in a capacitor is dependent on both its capacitance and the voltage applied across it, typically calculated using the formula U = ½CV². This concept is fundamental in understanding how devices can accumulate and then quickly discharge a large amount of energy when needed.

RC Circuit Discharge

An RC circuit is composed of a resistor and capacitor connected in series, and it describes how the capacitor discharges through the resistor over time. The discharge process is characterized by an exponential decay of voltage and current, governed by the time constant ? = RC. This model is critical in applications where a controlled pulse of current is necessary.

Ohm's Law

Ohm's Law is a basic principle in electrical circuits stating that the current through a resistor is proportional to the voltage across it and inversely proportional to its resistance (I = V/R). This straightforward relationship is essential for determining initial currents in situations where a voltage is suddenly applied across a resistive load.

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