Answer:
q = 0.036 C
Explanation:
Given that,
Current passes through a defibrillator, I = 18 A
Time, t = 2 ms
We need to find the charge moved during this time. We know that,
Electric current = charge/time
[tex]q=It[/tex]
Put all the values,
[tex]q=18\times 0.002\\\\q=0.036\ C[/tex]
So, 0.036 C of charge moves during this time.