Question

an electric circuit contains a battery that produces a voltage of 60 volts (V), a resistor with a resistance of 13 ohms, and an inductor with an inductance of 5 henrys (H), as shown in figure. Using calculus, it can be shown that the current I=I(t) (in amperes, A) t seconds after the switch is closed is I=60/13 (1-e^-13t/5). (A) use this equation to express the time t as a function of the current I. (B) after how many seconds is the current 2A?

Answer 1

@DanJS so how do you start solving It?

Answer 2

hi

Answer 3

hi

Answer 4

how u setup equation

Answer 5

@DanJS u there

Answer 6

yeah, ok

Answer 7

heres the question.. u see it

Answer 8

an electric circuit contains a battery that produces a voltage of 60 volts (V), a resistor with a resistance of 13 ohms, and an inductor with an inductance of 5 henrys (H), as shown in figure. Using calculus, it can be shown that the current I=I(t) (in amperes, A) t seconds after the switch is closed is I=60/13 (1-e^-13t/5). (A) use this equation to express the time t as a function of the current I. (B) after how many seconds is the current 2A?

Answer 9

it has to do with logaritms

Answer 10

it said using calculus

Answer 11

\[ I=\frac{ 60 }{ 13 } (1-e ^{\frac{ -13t }{ 5 }}).\]

Answer 12

yes

Answer 13

i do not know what they mean in part A

Answer 14

so it wants you to express time as a function of current, means solve for t

Answer 15

oh

Answer 16

take In of what

Answer 17

so all the initial explanation is not relevant

Answer 18

oh

Answer 19

first you want to isolate the e^(#) on one side by itself

Answer 20

i thought u divide the 60/13

Answer 21

first multiply both sides by 13/60

Answer 22

thats cancel i reckon

Answer 23

\[\frac{ 13 }{ 60 }*I = \frac{ 13 }{ 60 }* \frac{ 60 }{ 13 } * e ^{13t/5}\]

Answer 24

wheres the negative -13/5t

Answer 25

yeah its -13t/5 i forgot the neg sign

Answer 26

its: 13/60 I= e^-13t/5

Answer 27

So what you get now?

Answer 28

u take log of both sides

Answer 29

natural log

Answer 30

wait

Answer 31

forgot the 1

13/60 I = 1- e^-13t/5

Answer 32

u subtract 1 from both sides

Answer 33

add the e^# to both sides , then subtract, 13/60 I from both sides ...
you want the e^# to be positive

Answer 34

what happens to one

Answer 35

\[e ^{-13t/5} = 1 - \frac{ 13 }{ 60 }I\]

Answer 36

what happens to the one

Answer 37

take natural log of both sides

Answer 38

what u do now

Answer 39

yes

Answer 40

what happens to one on the In -13/60 I +1

Answer 41

\[-13t/5 = \ln(1-\frac{ 13 }{ 60 }I)\]

Answer 42

u trying to solve for t

Answer 43

you divide -13/5 to both sides

Answer 44

yes we want t = something

Answer 45

is this answer: In (-13/60I+1)/-13/5?

Answer 46

recall

\[\ln(a-b) = \ln \frac{ a }{ b }\]

Answer 47

yeah

Answer 48

im wrong

Answer 49

here

Answer 50

\[\frac{ 5 }{ -13 }*\frac{ -13 }{ 5 }t = \frac{ 5 }{ -13 }\ln(1 - \frac{ 13 }{ 60 }I)\]

Answer 51

\[t = \frac{ 5 }{ -13 }\ln(1 - \frac{ 13 }{ 60 }I)\]

Answer 52

that is part A,

Part B they tell you the current I = 2 amps

then you calculate t

Answer 53

now u plug in 2 for I

Answer 54

yeah

Answer 55

ok gn

Answer 56

\[t \approx0.218 s\]

Answer 57

good?