Find Voltage Across 47-uF Capacitor: A Math Homework Problem

[Basher]

Homework Statement

referring to the attatchment. the current consists of two semi circles. the question asks me to find the voltage across a 47-uF capacitor when t = 2ms.

Homework Equations

v(t) = 1/C ∫ i(t)dt +v(t0) {I realize their has to be limits for the integrand,I just can't type them}

A = [1/2.π.r^2]

The Attempt at a Solution

now after a while of thinking i realized i could find the area under the curve of this circle with the area formula multiplied by the capacitnce which gives me the right answer.

However, in my first attempt i assigned an equation to the first semi circle. this was

(2√(1 - (t - 1)^2))

then i integrated this by using trigonometric substitution.

I came up with this

arcsin(t - 1) + (t - 1)*(√(1 - (t - 1)^2))

i then evaluated at the upper limit t giving me the same formula.

then evaluated at t0 = 0 giving (-π/2). i then subtracted this away from the top formula.

so my entire formula is arcsin(t - 1) + (t - 1)*(√(1 - (t - 1)^2)) + (π/2).

substituting t = 2ms I'm left with (π/2) + (π/2) = π.

i multiply by 10^-6 because i have to account for the axes being both in ms and mA.

then i divide by 47-μF giving me double the correct answer. roughly 66.48mV. however the answer is half that. where did is screw up with the integral evaluation?

 

[tiny-tim]

Hi Basher!

(try using the X² button just above the Reply box )

(2√(1 - (t - 1)^2))

nooo … √(4 - (t - 1)²))

 

[Basher]

thanks. Did you have a crack at the problem?

 

[tiny-tim]

or a bash?

no … i was perfectly happy with …

now after a while of thinking i realized i could find the area under the curve of this circle with the area formula multiplied by the capacitnce which gives me the right answer.

 

[Basher]

hahaha. fair call. thanks