A 300 μF capacitor is charged to 9.0 V, then connected in parallel with a 5000 Ω resistor. The capacitor will discharge because the resistor provides a conducting pathway between the capacitor plates, but much more slowly than if the plates were connected by a wire. Let t=0 s be the instant the fully charged capacitor is first connected to the resistor. At what time has the capacitor voltage decreased by half, to 4.5 V?
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Solving Resistor Circuits
Problem 69
Textbook Question
A 150 μF defibrillator capacitor is charged to 1500 V. When fired through a patient’s chest, it loses 95% of its charge in 40 ms. What is the resistance of the patient’s chest?

1
Step 1: Recognize that the capacitor discharges through the patient's chest, which can be modeled as an RC circuit. The voltage across the capacitor during discharge follows the equation: , where is the time constant of the circuit.
Step 2: The time constant is given by the formula: , where is the resistance and is the capacitance.
Step 3: Use the information that the capacitor loses 95% of its charge in 40 ms. This means the voltage drops to 5% of its initial value. Substitute this into the discharge equation: , where and .
Step 4: Solve for the time constant using the equation: . Take the natural logarithm of both sides to isolate : . Rearrange to find : .
Step 5: Once is calculated, use the formula to solve for the resistance . Substitute the given capacitance and the calculated value of into the equation to find .

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Capacitance
Capacitance is the ability of a capacitor to store electrical charge per unit voltage. It is measured in farads (F), and in this case, the defibrillator capacitor has a capacitance of 150 μF (microfarads). The charge (Q) stored in a capacitor can be calculated using the formula Q = C × V, where C is capacitance and V is voltage.
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Capacitors & Capacitance (Intro)
Exponential Decay of Charge
When a capacitor discharges through a resistor, the charge decreases exponentially over time. The relationship is described by the equation Q(t) = Q0 * e^(-t/RC), where Q0 is the initial charge, R is resistance, and C is capacitance. In this scenario, the capacitor loses 95% of its charge in 40 ms, which can be used to find the resistance of the patient's chest.
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Amplitude Decay in an LRC Circuit
Ohm's Law
Ohm's Law relates voltage (V), current (I), and resistance (R) in an electrical circuit, expressed as V = I × R. In the context of the defibrillator, understanding this relationship is crucial for calculating the resistance of the patient's chest based on the current flowing through it as the capacitor discharges. This law helps in determining how the resistance affects the rate of charge loss.
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Resistance and Ohm's Law
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