Could someone help me with my homework please, i think i just need to know how to calculate the table of values so i can draw the graph... thanks
Exponential growth homework:
a. The formula for the instantaneous voltage Vi across a capacitor C being charged from 0 V up to the supply voltage E is given by:
Vi = E( 1 – e – t/CR )
where resistance R = 0.5 MΩ and e is the base of natural logs 2.718. If the supply voltage E = 10 V and the capacitor value C = 5.0 µF complete a table of 8 values of instantaneous voltage across the capacitor (Vi) against time t in (s) on x-axis from 0s to 4s.
b. Plot values of Vi (vertical) against time t (s) on a graph and draw a smooth curve through the points.
c. On your graph draw a tangent to the curve at t = 2 s. Draw the enclosing triangle and find and state the slope of the tangent. State the value of the rate of change of voltage in V/sec at t = 2 s.
d. Use differential calculus to find the value of the rate of change of voltage at
t = 2 s and analyse the result with that in c.
+129849
The equation is given by v = 10(1 - 2.718^[-t/(2.5)] )
Here's a graph of the function.......https://www.desmos.com/calculator/qgavq5phwq
(a) You should be able to handle the table part just by plugging in values 0, .5, 1, 1.5 .....4 for t
(b) This is already done, above....
(c).....I can see that the tangent line to the graph when t = 2 is going to intercept the t axis at a point where t < 0. It will be easier to answer (d) first......and then come back and draw this tangent line and the enclosing triangle........
(d) The derivative of the function can be found as follows :
v = 10(1 - 2.718^[-t/(2.5)] ) = 10 - 10(2.718)^(-.4t )
dv / dt = -10 ( -.4)ln(2.718)*(2.718)^(-.4t))
And evaluating this at t = 2 gives us a slope of about 1.797
And.....when t = 2, v = about 5.506
Now to find out where the tangent line intercepts the t axis, we can solve this equation.....
(5.506 - 0) / (2 - t) = 1.797
5.506 / 1/797 = 2 - t
t = 2 - 5.506/1/797 = about - 1.064
So....to draw the enclosing triangle .... draw a segment from (-1.064, 0) to (2,0).....then draw another segment from (2, 0) to (2, 5.506)....then draw a final segment from (2, 5.506) back to (-1.064, 0)And, of course the slope of the tangent at t = 2 = 1.797 = dv /dt at this point = the change in V/sec when t = 2
Lastly.....the equation of the tangent line can be found by:
v - 0 = 1.797(t - (-1.064))
v = 1.797t + 1.912
And here's the graph of the function and the tangent line to the graph when t = 2
+223 Maybe I´m not qualified for answering your question and maybe I don´t understand english properly so don´t trust me completly
...but isn´t there just to put the numbers into the formula and use different values for t?
8 values means 8 different times. as a suggestion every half second to start with.
+394 The RC - Time Constant for this circuit is (0.5e6)(5e-6) = 2.5 seconds.
Use that as the basis for the graph.
At 5*2.5 = 12.5seconds the capacitor is 99.333 charged.
Make 16 Tic marks on the graph from 0 to 12.5 at 0.78s intervals (also mark the 2 second interval).
_7UP_
+129849
The equation is given by v = 10(1 - 2.718^[-t/(2.5)] )
Here's a graph of the function.......https://www.desmos.com/calculator/qgavq5phwq
(a) You should be able to handle the table part just by plugging in values 0, .5, 1, 1.5 .....4 for t
(b) This is already done, above....
(c).....I can see that the tangent line to the graph when t = 2 is going to intercept the t axis at a point where t < 0. It will be easier to answer (d) first......and then come back and draw this tangent line and the enclosing triangle........
(d) The derivative of the function can be found as follows :
v = 10(1 - 2.718^[-t/(2.5)] ) = 10 - 10(2.718)^(-.4t )
dv / dt = -10 ( -.4)ln(2.718)*(2.718)^(-.4t))
And evaluating this at t = 2 gives us a slope of about 1.797
And.....when t = 2, v = about 5.506
Now to find out where the tangent line intercepts the t axis, we can solve this equation.....
(5.506 - 0) / (2 - t) = 1.797
5.506 / 1/797 = 2 - t
t = 2 - 5.506/1/797 = about - 1.064
So....to draw the enclosing triangle .... draw a segment from (-1.064, 0) to (2,0).....then draw another segment from (2, 0) to (2, 5.506)....then draw a final segment from (2, 5.506) back to (-1.064, 0)And, of course the slope of the tangent at t = 2 = 1.797 = dv /dt at this point = the change in V/sec when t = 2
Lastly.....the equation of the tangent line can be found by:
v - 0 = 1.797(t - (-1.064))
v = 1.797t + 1.912
And here's the graph of the function and the tangent line to the graph when t = 2
