Network Analysis

Tutorial on Inductors and Capacitors - Part 2

That's right.
v(t) = 3 + 6t² + 5e^-4t + 2sin(8t) volts, so
v(0) = 3 + 6x0 + 5x1 + 2x0 = 8V
resistor current = v(0)/R = 8/10 = 0.8A

Let's show all the branch currents on the circuit diagram:

We now know that i_R(0) = 0.8A.

In general,

i_R(t) = v(t)/R = 0.3 + 0.6t² + 0.5e^-4t + Xsin(8t) amps

Find X.

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