The current i in a certain circuit is given by
i=6.25e^-125t (amperes)
where t is the time in seconds. At what time will the current be 1.00 A?
To find the time at which the current will be 1.00 A, we need to solve the equation i = 1.00 for t.
Given: i = 6.25e^(-125t)
Setting i = 1.00, we have:
1.00 = 6.25e^(-125t)
Taking the natural logarithm of both sides of the equation, we get:
ln(1.00) = ln(6.25e^(-125t))
Using the logarithmic property ln(e^x) = x, we simplify the equation to:
0 = ln(6.25) - 125t
Simplifying further, we have:
-125t = ln(6.25)
Dividing both sides by -125, we find:
t = (ln(6.25)) / (-125)
Calculating this expression using a calculator, we get:
t ≈ -0.022 seconds
Note that the negative sign indicates that t is a time in the past. In this case, the current of 1.00 A occurred approximately 0.022 seconds ago.
To find the time at which the current is 1.00 A, we need to solve the equation i = 1.00 A and find the corresponding value of t.
Given the equation: i = 6.25e^(-125t)
Substituting i = 1.00 A, we get:
1.00 A = 6.25e^(-125t)
To isolate t, we need to divide both sides by 6.25:
1.00 A / 6.25 = e^(-125t)
0.16 = e^(-125t)
To eliminate the exponential term, we need to take the natural logarithm of both sides:
ln(0.16) = ln(e^(-125t))
Using the property ln(e^x) = x, we simplify:
ln(0.16) = -125t
Now we can solve for t by dividing both sides by -125:
t = ln(0.16) / -125
Using a calculator, we find:
t ≈ -0.2668 / -125
t ≈ 0.002134
Therefore, the time at which the current will be 1.00 A is approximately 0.002134 seconds.
1/6.25 = e^-(125 t)
.16 = e^-(125 t)
ln .16 = -125 t
-1.83 = -125 t