# Problem: A heart defibrillator being used on a patient has an RC time constant of 10.0 ms due to the resistance of the patient and the capacitance of the defibrillator.(a) If the defibrillator has an 8.00 F capacitance, what is the resistance of the path through the patient?(b) If the initial voltage is 12.0 kV, how long does it take to decline to 6.00 ×102  V?

###### FREE Expert Solution

The time constant for the RC circuit:

$\overline{){\mathbf{\tau }}{\mathbf{=}}{\mathbf{R}}{\mathbf{C}}}$

The voltage across the capacitor when discharging:

$\overline{){{\mathbf{V}}}_{{\mathbf{C}}}{\mathbf{=}}{\mathbf{\epsilon }}{{\mathbf{e}}}^{\frac{\mathbf{-}\mathbf{t}}{\mathbf{\tau }}}}$

(a)

Resistance, R = τ/C

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###### Problem Details

A heart defibrillator being used on a patient has an RC time constant of 10.0 ms due to the resistance of the patient and the capacitance of the defibrillator.

(a) If the defibrillator has an 8.00 F capacitance, what is the resistance of the path through the patient?

(b) If the initial voltage is 12.0 kV, how long does it take to decline to 6.00 ×102  V?