Question

During surgery, a current as small as 20.0 μA applied directly to the heart may cause ventricular fibrillation. If the resistance of the exposed heart is 300Ω, what is the smallest voltage that poses this danger?

a) 6.00 mV
b) 60.0 mV
c) 0.60 V
d) 0.00600 V

Answer 1

The smallest voltage that can cause ventricular fibrillation with a current of 20.0 μA and a resistance of 300Ω is 6.00 mV. This implies that special electrical safety precautions are necessary during surgery.

Step-by-step explanation:

To determine the smallest voltage that can cause ventricular fibrillation when a current of 20.0 μA is applied to the heart, we can use Ohm's Law, which states V = IR, where V is voltage, I is current, and R is resistance. Plugging in the values provided:

• I = 20.0 μA (which is 20.0 x 10-6 A)
• R = 300Ω

Then:

V = (20.0 x 10-6 A) × 300Ω = 6.00 x 10-3 V

This voltage corresponds to 6.00 mV, which is the smallest voltage that could pose a danger of ventricular fibrillation to the heart during surgery. This information indicates that special electrical safety precautions are indeed needed during surgical procedures to prevent such a small voltage from being applied to the heart.