Electronic – First order circuit with t=0+ , t=0-

It's hard to tell from the photo, so I may be incorrect, but it appears that the "G" pin on your PUT is connected to the breadboard, but that nothing is connected to a rail that it shares. It may sound like the obvious, but it helps to double check that everything is firmly connected to the board and that the board is being powered reliably (I.E. No dead batteries, etc.). More often than not, it's simple mistakes that can make a circuit not work no matter what you try. It won't hurt to double check that you have your capacitor and LED polarities correct too.

Laplace Transform:

Solving in the s-domain will be easier. Writing the current division formula,

$$i_c (s) = \frac{R}{R+1/Cs}\times I(s)$$ $$i_c (s) = \frac{1}{1+1/5s}\times \frac{10}{s} =\frac{10}{s+0.2}$$

Taking the Laplace inverse,

$$i_c(t) = 10e^{-0.2t}u(t)$$ $$\therefore i_c(0^+) = 10A$$

$$------------------------------------$$

Another view to solve this problem: Capacitor acts as a short circuit for high frequency currents. At t = 0, a sudden switching from 0 to 10A (high frequency) occurs and capacitor acts as short circuit and hence the entire current will be flowing through capacitor. Or, \$i_c(0^+) = 10A\$.

$$------------------------------------$$ Using continuity condition on the capacitor's voltage:

Capacitor at \$t = 0^+\$ can be replaced with a voltage source with voltage = \$v_c(t=0^-)\$. But \$v_c(t=0^-) = 0\$. Voltage source with voltage = 0 is short circuit. Hence the entire current will be flowing through capacitor. Or, \$i_c(0^+) = 10A\$.