Question

A heart defibrillator being used on a patient has an RC time constant of 17.0 ms due to the resistance of the patient and the capacitance of the defibrillator. If the defibrillator has a 9.75 µF capacitance, what is the resistance of the path through the patient?

a) 1.75 kΩ
b) 5.29 kΩ
c) 11.11 kΩ
d) 20.58 kΩ

Answer 1

The resistance of the path through a patient with an RC time constant of 17.0 ms and a defibrillator capacitance of 9.75 µF is approximately 1.75 kΩ.

Step-by-step explanation:

The subject of this question is Physics, specifically dealing with the concept of an RC time constant within the context of a heart defibrillator. The key components here are resistance, capacitance, and their relationship in determining the RC time constant. To calculate the resistance of the path through the patient, given the RC time constant and the capacitance of the defibrillator, we use the formula τ = RC, where τ (tau) is the time constant, R is the resistance, and C is the capacitance.

Given:

• RC time constant (τ) = 17.0 ms = 17.0 × 10⁻³ s
• Capacitance (C) = 9.75 µF = 9.75 × 10⁻¶ F

We can rearrange the formula to solve for R:

R = τ / C

R = (17.0 × 10⁻³) / (9.75 × 10⁻¶) = 1.7436 × 10⁳ Ω = 1.74 kΩ

Therefore, the resistance of the path through the patient is approximately 1.74 kΩ, which when rounded to two significant figures becomes 1.75 kΩ (option a).