Question

The voltage across a 8.5-F capacitor is 2te–3t V. Find the current and the power.

Answer 1

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