Network Analysis

Tutorial on Inductors and Capacitors - Part 3

That's right.
i_R(t) = v(t)/R
i_R(t) = [3 + 6t² + 5e^-4t + 2sin(8t)]/10
i_R(t) = 0.3 + 0.6t² + 0.5e^-4t + 0.2sin(8t) amps

Now let's try to find i_C(t) .

Recall the general formula for calculating the current in a capacitor:
i_C(t) = C(dv_C/dt)

Notice that the voltage across the capacitor is the source voltage, v(t). This gives us the following:
i_C(t) = 20x10^-6[dv(t)/dt]

It appears that we must calculate the time derivative of v(t):
dv(t)/dt = d[3 + 6t² + 5e^-4t + 2sin(8t)]/dt

Let's do this one term at a time. What's the derivative of the 3?

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