Question

Circuit analysis with calculus.

Answer 1

@Loser66

Answer 2

I am helpless here

Answer 3

but it's calculus

Answer 4

The first step is to find the charge on the capacitor after 500 micro-seconds:
\[\large Q=\int\limits_{t=0}^{t=500}50e^{-2000t}.dt\ millimicrocoulombs\]
The next step is to find the voltage across the capacitor after 500 micro-seconds using the formula:
\[\large V=\frac{Q}{C}\]
Finally the energy stored in the capacitor at t = 500 micro-seconds can be found from:
\[\large W=\frac{CV^{2}}{2}\ joules\]

Answer 5

thank you for helping me

Answer 6

what take the integral? I thought the derivative? So the area under a function of current respect to time is joules?

Answer 7

Q=di/dt?

Answer 8

before we proceed can we discuss units and how everything relates?

Answer 9

So the last equation is the capacitor energy equation

Answer 10

\[\large Q=\int\limits_{t=a}^{t=b}i.dt\]

Answer 11

so charge=di/dt?

Answer 12

current respect to time?

Answer 13

so i get .015803mA

Answer 14

after integration

Answer 15

i plugged the integral in my calculator with my limits of integration to be 0 to 500E-6 and got .015803

Answer 16

so?

Answer 17

Please wait a little while, I have had to take a phone call.

Answer 18

okay thanks for letting me know

Answer 19

The charge in coulombs = amperes * seconds. Therefore the value of the definite integral is:
\[\large 0.0158\times10^{-3}\ coulombs\]
the reason being that the current is stated in mA.

Now to find the voltage across the capacitor after 500 microseconds we use:
\[\large V=\frac{Q}{C}=\frac{0.0158\times10^{-3}}{0.5\times10^{-6}}\]

Answer 20

okay I agree now.

Answer 21

where did you get C?

Answer 22

C is the 0.5uF capacitor shown in the question.

Answer 23

okay, it's been a long day...

Answer 24

v=31.606v

Answer 25

then I use the equation w=1/2(CV^2)

Answer 26

I am not getting the corret answer.