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

In summary, the student attempted to find the voltage across a 47-uF capacitor when t = 2ms by using the area formula and the trigonometric substitution. However, they made a mistake with the integral evaluation and ended up with the wrong answer.
  • #1
Basher
13
0

Homework Statement



IMG.jpg


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?
 
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  • #2
Hi Basher! :smile:

(try using the X2 button just above the Reply box :wink:)
Basher said:
(2√(1 - (t - 1)^2))

nooo … √(4 - (t - 1)2)) :wink:
 
  • #3
thanks. Did you have a crack at the problem?
 
  • #4
or a bash? :wink:

no … i was perfectly happy with …
Basher said:
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.

:smile:
 
  • #5
hahaha. fair call. thanks
 

What is the purpose of finding the voltage across a 47-uF capacitor?

The purpose of finding the voltage across a capacitor is to determine the potential difference between the two terminals of the capacitor. This value is important in understanding the behavior and functionality of the capacitor in a circuit.

How do you calculate the voltage across a 47-uF capacitor?

The voltage across a capacitor can be calculated using the formula V = Q/C, where V is the voltage, Q is the charge stored in the capacitor, and C is the capacitance. In this case, the capacitance is given as 47-uF.

Why is finding the voltage across a capacitor important in circuit analysis?

Finding the voltage across a capacitor is important in circuit analysis because it helps in determining the energy stored in the capacitor and the voltage drop across the capacitor. This information is crucial in understanding the overall behavior of the circuit.

What units are used for the voltage across a capacitor?

The voltage across a capacitor is typically measured in volts (V) in the SI unit system. However, it can also be expressed in other units such as millivolts (mV) or kilovolts (kV).

How does the voltage across a capacitor change with time?

The voltage across a capacitor changes over time as the capacitor charges or discharges. Initially, when the capacitor is fully discharged, the voltage across it is zero. As it charges, the voltage gradually increases until it reaches the same value as the applied voltage. When the capacitor discharges, the voltage decreases until it reaches zero again.

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