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.

• If the time constant is 20nS,rise time is 44nS. You require a CRo with rise time of 5nS and bandwidth of a scope of that type will be 80 to 100 MHz. You can see in CRO provided, you take adequate care in layout etc so that what you see is what it is.