Answer:
Explanation:
Given the capacitance of a capacitor = 8.5F
Voltage across the capacitor = 2te^-3tV
The current across a capacitor is expresses as shown;
I = Cdv/dt
Given V = 2te^-3tV
dv/dt = 2t(-3e^-3t) + 2(e^-3t)
dv/dt = -6te^-3t + 2e^-3t
Factor out the common term
dv/dt = -2e^-3t{3t-1}
If C = 8.5F
I = 8.5 × -2e^-3t{3t-1}
I = -17.0e^-3t{3t-1} A
Power is the product of current and voltage.
P = IV
Given I = -17.0e^-3t{3t-1} A
V = 2te^-3t V
Power = -17.0e^-3t{3t-1} × 2te^-3t
Power = (17e^-3t - 51te^-3t) × 2te^-3t
Power = 34te^{-3t-3t} - 102t²e^{-3t-3t}
Power = 34te^-6t - 102t²e^-6t
Power = 34te^-6t{1-3t} Watts
