Calculating the average voltage?

Question

find the average value of V=100e^-((10^3)t) between the instant v=9 to instant v=1 the answer is 3.642v

pls show me how the answer ended up to be 3.642v

Answer 1

I am not sure, but I think if you take the integral of the equation (as is the area under the curve) for that region you will get a good start.

Then maybe divide by the number of seconds in the region.

Answer 2

I have answered this question 2 times and my answers disappeared.

An avg of a function between a defined interval is the integral of that function over that interval divided by that interval.

It seems to me that you don't want to put effort to get the answer. Look at below for the step by step calculation:

We need to find corresponding t values for V = 9 and V = 1, Let say t9 and t1

for v = 9:

  9 = 100*e^(-10^3t)

   t = - (ln 0.09) ms

   t = 2.4084 ms --> [let call this is t9]

for v = 1:

  1 = 100*e^(-10^3t)

   t = - (ln 0.01) ms

     = 4.606 ms--> [let say this is t1]

Avg of V = 1/(t1-t9)ms* integral of (100*e^(-1000t)) dt

Avg of V = 1000/(4.606 -2.4084)* integral of (100*e^(-1000t)) dt

Avg of V = 455*(-1/10 *e^(-1000t)) <-- plug in t9 and t1 value

Avg of V = 455*(0.008) = 3.64 v

Answer 3

i can solve it, i know it's parallel circuit but where is the resistor and current?

show me the amount of resistor and current because

resistor x current = volts