January 1, 2013 at 5:35 pm #21121
You can approximate this as an RC circuit. http://en.wikipedia.org/wiki/RC_circuit
If 12 V draws 4 A, then you could pretend that it’s a 3 ohm resistive load.
Then V(t) = V_0 exp(-t/RC), or V(t) / V_0 = exp(-t/RC)
Taking the natural log of both sides,
ln(V(t)/V_0) = -t/RC
For your case, we are solving for t=0.5 s. V(t = 0) = 12 V, and V(t = 0.5 s) = 10 V.
Then, ln(10/12) = -0.5 s/RC
Approximately given by -.182 = -0.5 s/RC, or RC = -0.5 s/ -.182 = 2.74
Putting back in R = 3 ohms, that gives C = 2.74/3, or about 0.9 F.
So, you’ll need roughly a 1 F capacitor rated for at least 12 V, or in practice 1.5 X higher just in case the actual voltage goes higher. It’s actually common to find capacitors in this range (1-5 F, 16-24 V rating), manufactured for use in car stereos, in the range $20-$100.