Analysis for Challenge 11

If Node X remains at 0V, V1 will be forcing current through the capacitor from left-to-right, such that

               i C1 = C1dV/dt

               i C1 = 1F•5V/s

               i C1 = 5A

If Node X remains at 0V, V3 will be forcing current through the inductor from right-to-left, so that

               vL1 = L1di/dt

               5V = 1H(di/dt)

            di/dt = 5A/s

Then, if Node X is to remain at 0V, V2 must necessarily force Node Y negative in such a way that i R1 is given by

               i R1 = u(t)[5A + (5A/s)t]

Where u(t) is the unit step function, defined as

             u(t) = 0 for t < 0

             u(t) = 1 for t ≥ 0

So, if R1 is 1Ω, we can write, for vY

               vY = 0 – u(t)[5A + (5A/s)t]•1Ω

               vY = 0 – u(t)(1 + t)5V

And this, then, is the voltage at Node Y that forces node X to be 0V (a virtual ground) for the duration of the event.