Anonymous
Anonymous asked in Science & MathematicsEngineering · 6 years ago

# How do you find peak current?

Given:

L = 140 pF

R = 140 ohms

v = 18300 V

Problem is along the lines of; person shuffling feet along dry carpet acts as a charged capacitor with 140 pF. The capacitance my be charged to 18300V. When the person touches a water faucet they experience a light shock, finger acts as a resistor 140 ohms.

Find peak current.

p.s. I found out the peak current is NOT -502A or 502A

I have found the time constant to be 1.96 * 10^-8 sec

Thanks!

Relevance

It's more obvious than you think (or I haven't woken up enough yet.) It's just 18300V ⁄ 140Ω ≈ 130.7A. That's so, because the capacitor is charged up and acts like a simple voltage source (for only a moment) that discharges through the 140Ω resistor.

But let's do it more properly. You know the equation for current in a capacitor as:

1.      I = C ⋅ dV ⁄ dt

But what is dV ⁄ dt? Well, in a discharging RC circuit loop, you know this:

2.      V(t) = V₀ ⋅ e^(-t ⁄ (R⋅C))

Where V₀ = 18300V, C = 140pF, and R = 140Ω, in this case. Taking the derivative with respect to time, t, you get:

3.      d( V(t) ) = d( V₀ ⋅ e^(-t ⁄ (R⋅C)) )

4.      d( V(t) ) = V₀ ⋅ d( e^(-t ⁄ (R⋅C)) )

5.      d( V(t) ) = V₀ ⋅ e^(-t ⁄ (R⋅C)) ⋅ d( -t ⁄ (R⋅C) )

6.      d( V(t) ) = V₀ ⋅ e^(-t ⁄ (R⋅C)) ⋅ -1 ⁄ (R⋅C) d( t )

7.      d V(t) ⁄ dt = -V₀ ⁄ (R⋅C) ⋅ e^(-t ⁄ (R⋅C))

This will be at its fastest (peak) when t = 0, so:

8.      d V(t) ⁄ dt = -V₀ ⁄ (R⋅C), peak value

Substituting the magnitude of (8) into (1):

9.      I = C ⋅ V₀ ⁄ (R⋅C) = V₀ ⁄ R

Which gets back to what was computed at the start. You don't need the time constant.

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