Analysis for Challenge 6

James’ Circuit

L2, L3, L4, and L5 have a combined parallel inductance of 1μH (equivalent to L1).  Thus, half the input current will always be flowing in L1, and we can write

                   i L1 = t2/2

Then, we solve for t when i L1 is 1 Ampere, as follows:

                      1 = t2/2

                     t2 = 2

                       t = √2 seconds

The voltage on L1 is found using the following relationship:

                    vL1 = L1di/dt

For L1, di/dt = t Amperes/sec.  Therefore we can write

                    vL1 = L1t(A/sec)

In this case t=√2.  Thus,

                    vL1 = 1μH(√2)(A/sec)

                    vL1 = (√2)μV   ←  voltage on L1 when the current in L1 is 1 Ampere

Lisa’s Circuit

C2, C3, C4, and C5 have a combined series capacitance of 1μF (equivalent to C1).  Thus, half the input voltage will always appear on C1, and we can write

                   vC1 = t2/2

Then, we solve for t when vC1=1 Volt

                      1 = t2/2

                     t2 = 2

                       t = √2 seconds

The current in C1 is found using the following relationship:

                    iC1 = C1dV/dt

For C1, dV/dt = t volts/sec.  Therefore we can write

                    iC1 = C1t(V/sec)

In this case t=√2.  Thus,

                    iC1 = 1uF(√2)(V/sec)

                    iC1 = (√2)μA   ←  current in C1 when the voltage on C1 is 1V

Did You Notice?

Both input sources have finite slopes at t=0.

  •  If i had instantly stepped from 0 to some non-zero value, Ldi/dt=∞
  •  If v had instantly stepped from 0 to some non-zero value, Cdv/dt=∞