Electric Current

There are two approaches to this problem. One is more sophisticated than the other. Let's use do the basic method first.

1. Use the definition of current and solve for charge.

   ∆q = I∆t

   ⇐

   I =  ∆q ∆t

   Use the values for one pulse and multiply by the number of pulses.

   ∆q = nI∆t ∆q = 95(3 A)(80 × 10−6 s) ∆q = 0.0228 C

   This is considered (for lack of more precise words) a lot of charge.

2. Now use the defintion of current.

   I =	∆q
   	∆t
   I =	0.0228 C
   	5 s
   I = 0.00456 A

   This is not a lot of current. You need something between 70 and 100 milliamps to die by electrocution. This is not even 5 mA.

Both these answers are twice too big. The numbers for this problem came from a report by the US Department of Justice. The report cited a claim that this particular "conducted energy device" delivers about 100 microcoulombs of charge per pulse and 2.1 milliamps of average current. The discrepency arises because the current in one pulse is not constant. Imagine a triangle with a base of 80 µs and a height of 3 A. The current starts at zero, hits a peak, and returns to zero. The total charge delivered is the area under this triangle.

1. Compute the charge delivered in one pulse. I'm not going to use scientific notation for this solution. I prefer using SI prefixes.

   ∆qpulse = ½bh ∆qpulse = ½(80 µs)(3 A) ∆qpulse = 120 µC

   This agrees with the report. It's about 100 microcoulombs. Mutiply this by 95 pulses to get the total charge of one cycle.

   ∆qtotal = 95(120 µC) ∆qtotal = 11,400 µC ∆qtotal = 11.4 mC

   This is still considered a lot of charge, which is why these weapons can produce intolerable pain (and why they have been associated with torture).

2. Back to the defintion of current.

   I =	∆q
   	∆t
   I =	11.4 mC
   	5 s
   I = 2.28 mA

   This is similar to the value found in the report. It is still not a lot of current, which is why these weapons rarely kill.

Answer it.

Answer it.

Answer it.