A cauterizer, used to stop bleeding in surgery, puts out 2.30 mA at 14.0 kV. What is its power output?
What is the resistance of the path?
power=Voltage*current=14E3*2.3E-3 watts
I got the first part of the question right.
It was 32.2 W
How would I do the second part?
To find the power output of the cauterizer, we can use the formula:
Power (P) = Current (I) * Voltage (V)
Given that the current is 2.30 mA (or 0.00230 A) and the voltage is 14.0 kV (or 14,000 V), we can plug these values into the formula:
P = 0.00230 A * 14,000 V
Calculating this will give us the power output of the cauterizer.
Now, to find the resistance of the path, we can use Ohm's Law:
Resistance (R) = Voltage (V) / Current (I)
Given that the voltage is 14.0 kV (or 14,000 V) and the current is 2.30 mA (or 0.00230 A), we can plug these values into the formula:
R = 14,000 V / 0.00230 A
Calculating this will give us the resistance of the path.