Question

part 1 of 3
my score: 31.77/34 pts (88.43%)
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the equation governing the amount of current i (in amperes) (in seconds) after t seconds in a simple rl circuit consisting of a resistance r (in ohms) and an inductance l (in henrys) and an electromotive force e (in volts) is i = e/r 1 - e^(-rt/l).
(a) if e = 17 volts, r = 13 ohms, and l = 1 henrys, how long does it take to obtain a current of 0.5 ampere?
(round to four decimal places as needed.)
t = (seconds)

Explanation:

Step1: Rearrange the given formula

The formula for the current in an RL - circuit is $I=\frac{E}{R}(1 - e^{-\frac{Rt}{L}})$. We want to find $t$ when $I = 0.5$ amperes, $E = 17$ volts, $R=13$ ohms and $L = 1$ henry. First, substitute the given values into the formula: $0.5=\frac{17}{13}(1 - e^{-\frac{13t}{1}})$.

Step2: Simplify the equation

First, simplify $\frac{17}{13}(1 - e^{- 13t})=0.5$. Then, $1 - e^{-13t}=\frac{0.5\times13}{17}=\frac{6.5}{17}\approx0.3824$.

Step3: Isolate the exponential term

We get $e^{-13t}=1 - 0.3824 = 0.6176$.

Step4: Take the natural - logarithm of both sides

Taking the natural logarithm of both sides, $\ln(e^{-13t})=\ln(0.6176)$. Since $\ln(e^{-13t})=-13t$, we have $-13t=\ln(0.6176)$.

Step5: Solve for $t$

$t=-\frac{\ln(0.6176)}{13}$. We know that $\ln(0.6176)\approx - 0.4805$. So $t=\frac{0.4805}{13}\approx0.037$.

Answer:

$0.037$