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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.

 May 16, 2015

Best Answer 

 #3
avatar+128085 
+10

 

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

 

 

 

 May 17, 2015
 #1
avatar+223 
+5

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.

 May 16, 2015
 #2
avatar+394 
+5

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_

 May 16, 2015
 #3
avatar+128085 
+10
Best Answer

 

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

 

 

 

CPhill May 17, 2015

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