Ch 28: Fundamentals of Circuits

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 28: Fundamentals of Circuits

Problem 69

Chapter 28, Problem 69

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?

Verified step by step guidance

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:

V_{t} = V_{0} e^{- \frac{t}{τ}}

, where

τ

is the time constant of the circuit.

Step 2: The time constant

τ

is given by the formula:

τ = R C

, where

R

is the resistance and

C

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:

V_{t} = V_{0} e^{- \frac{t}{τ}}

, where

V_{t} = 0.05 V_{0}

and

t = 40 ms

Step 4: Solve for the time constant

τ

using the equation:

e^{- \frac{t}{τ}} = 0.05

. Take the natural logarithm of both sides to isolate

\frac{t}{τ}

- \frac{t}{τ} = \ln (0.05)

. Rearrange to find

τ

τ = \frac{t}{-}

Step 5: Once

τ

is calculated, use the formula

τ = R C

to solve for the resistance

R

. Substitute the given capacitance

C = 150 μ F

and the calculated value of

τ

into the equation to find

R

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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.

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.

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

Related Practice

Textbook Question

The capacitor in an RC circuit is discharged with a time constant of 10 ms. At what time after the discharge begins are (a) the charge on the capacitor reduced to half its initial value and (b) the energy stored in the capacitor reduced to half its initial value?

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Textbook Question

A 15 μF capacitor charged to 12 V is discharged through a resistor. The energy stored in the capacitor decreases by 50% in 0.25 s. What is the value of the resistance?

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Textbook Question

How much current flows through the bottom wire in FIGURE P28.66, and in which direction?

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Textbook Question

The capacitor in FIGURE P28.73 begins to charge after the switch closes at t = 0 s. Find an expression for the current I at time t. Graph I from t = 0 to t = 5τ.

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Textbook Question

The flash on a compact camera stores energy in a 120 μF capacitor that is charged to 220 V. When the flash is fired, the capacitor is quickly discharged through a lightbulb with 5.0 Ω of resistance. At what rate is the lightbulb dissipating energy 250 μs after the flash is fired?

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Textbook Question

A circuit you’re using discharges a 20 μF capacitor through an unknown resistor. After charging the capacitor, you close a switch at t = 0 s and then monitor the resistor current with an ammeter. Your data are as follows: Use an appropriate graph of the data to determine (a) the resistance and (b) the initial capacitor voltage.

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