A 313 μF defibrillator capacitor is charged to 4.94×103 V. When fired through a patients chest, it loses 95 % of its charge in 40 ms.

What is the resistance of the patients chest?

somebdy help

See prev post

To calculate the resistance of the patient's chest, we need to utilize Ohm's Law, which states that resistance (R) is equal to voltage (V) divided by current (I). In this case, the voltage is given, but we need to find the current.

To find the current, we can use the principle of conservation of charge, which states that the charge remains constant before and after the discharge. We know that the capacitor loses 95% of its charge in 40 ms. Therefore, the remaining charge is 5% of the initial charge.

To find the remaining charge on the capacitor, we can use the formula:

Q = C * V

Where Q is the charge, C is the capacitance, and V is the voltage. Plugging in the given values:

Q = (313 × 10^-6 F) * (4.94 × 10^3 V)
= 1.546 × 10^-3 C

Since the remaining charge is 5% of the initial charge, we can calculate the remaining charge:

Remaining Charge = 0.05 * 1.546 × 10^-3 C
= 7.73 × 10^-5 C

Now, we can find the current (I) by dividing the remaining charge by the time (t):

I = Remaining Charge / t
= (7.73 × 10^-5 C) / (40 × 10^-3 s)
= 1.933 A

Finally, we can calculate the resistance (R) by dividing the voltage (V) by the current (I):

R = V / I
= (4.94 × 10^3 V) / (1.933 A)
= 2551 Ω

Therefore, the resistance of the patient's chest is approximately 2551 Ω.