Question #6281

i(t)=(V/R) * (1-e^-(R/L)(t))
(a) Define an anonymous function answerE that returns i(t) given input t.
(b) The final value for the current is 2 amps. Find a time tm when the
current is at least 99% of the nal value but strictly less than 2. Store
this value in the variable answerF.
(c) Create a vector in answerG using linspace with ten evenly spaced
elements where 0 as the first element and tm as the last element.
(d) Create a vector in answerH which contains the current at each time in
vector answerG.

Expert's answer

a) To define an anonymous function answerE that returns i(t) given input t run the command:

answerE=@(t)(V/R[/url])*( 1-exp(-(R*t/L)) );

b) We have that the final value for the current is 2 amps.

This means that V/R = 2.

In order to find tm we need to know the value of R/L.

In other words we should give some values for V, R, L.

So set some values and then redefine the function:

V=2;

R=1;

L=0.5;

answerE=@(t)(V/R[/url])*( 1-exp(-(R*t/L)) );

To find a time tm when the current is at least 99% of the final value but strictly less than 2,

and store this value in the variable answerF, run

answerF = fsolve(@(t)answerE(t)-0.99*V/R,1)

we will get the result:

>> answerF = 2.3026

(c) To create a vector in answerG using linspace with ten evenly spaced elements where 0 as the first

element and tm as

the last element run

answerG=linspace(0,answerF,10)

which will give the output

>> answerG =

>> 0.00000 0.25584 0.51169 0.76753 1.02337 1.27921 1.53506 1.79090 2.04674 2.30259

(d) To create a vector in answerH which contains the current at each time in vector answerG

run

answerH=answerE(answerG)

which will give the output

>> answerH =

>> 0.00000 0.80103 1.28124 1.56911 1.74169 1.84515 1.90717 1.94435 1.96664 1.98000

answerE=@(t)(V/R[/url])*( 1-exp(-(R*t/L)) );

b) We have that the final value for the current is 2 amps.

This means that V/R = 2.

In order to find tm we need to know the value of R/L.

In other words we should give some values for V, R, L.

So set some values and then redefine the function:

V=2;

R=1;

L=0.5;

answerE=@(t)(V/R[/url])*( 1-exp(-(R*t/L)) );

To find a time tm when the current is at least 99% of the final value but strictly less than 2,

and store this value in the variable answerF, run

answerF = fsolve(@(t)answerE(t)-0.99*V/R,1)

we will get the result:

>> answerF = 2.3026

(c) To create a vector in answerG using linspace with ten evenly spaced elements where 0 as the first

element and tm as

the last element run

answerG=linspace(0,answerF,10)

which will give the output

>> answerG =

>> 0.00000 0.25584 0.51169 0.76753 1.02337 1.27921 1.53506 1.79090 2.04674 2.30259

(d) To create a vector in answerH which contains the current at each time in vector answerG

run

answerH=answerE(answerG)

which will give the output

>> answerH =

>> 0.00000 0.80103 1.28124 1.56911 1.74169 1.84515 1.90717 1.94435 1.96664 1.98000

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