Rl series circuit help!!!?
i am currently working on a question for my class i need to find the time when the current reaches 3amps the equation is: i=3.75(1-e^-t/0.5). So if you replaced the i with 3, how would you rearrange the equation to solve for t?
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i=3.75(1- e^-t/0.5)
i/3.75= 1- e^-t/0.5
i/3/75 - 1 = -e^-t/0.5
note the negative sign in the exponential "e", just send it to the other side
-i/3.75 + 1 = e^-t/0.5
e^-t/0.5 = 1 - i/3.75
-t/0.5 = ln ( 1 - i/3.75)
-t = 0.5 ln (1 - i/3.75)
t= -0.5 ln (1 - i/3.75) replace i by 3 and you get the value of t
1 - e^(-t/0.5) = 0.8
- e^(-t/0.5) = 0.8-1 = -0.2
e^(-t/0.5) = 0.2
-t/0.5 = ln(0.2)
-t = 0.5ln(0.2)
t = -0.5ln(0.2)
t = -0.5 * -1.609
t = 0.8047 seconds