Question

question part 2 of 3
5.4 hw - logarithmic functions
the equation governing the amount of current i (in amperes) after time t (in seconds) in a simple rl circuit consisting of a resistance r (in ohms), an inductance l (in henrys) and an electromotive force e (in volts) is (i=\frac{e}{r}(1 - e^{-\frac{rt}{l}})).
(a) if (e = 17) volts, (r = 13) ohms, and (l = 1) henry, how long does it take to obtain a current of 0.5 ampere? (round to four decimal places as needed)
(t=)0.0371 seconds
(b) how long does it take to obtain a current of 1.0 ampere? (round to four decimal places as needed)
(t=)

Explanation:

Step1: Recall RL - circuit current formula

The formula for the current in an RL - circuit is $I(t)=\frac{E}{R}(1 - e^{-\frac{Rt}{L}})$. We can re - arrange it to solve for $t$ when given $I$, $E$, $R$, and $L$. First, rewrite the formula as $1 - e^{-\frac{Rt}{L}}=\frac{I R}{E}$, then $e^{-\frac{Rt}{L}}=1-\frac{I R}{E}$, and taking the natural logarithm of both sides gives $-\frac{Rt}{L}=\ln(1 - \frac{I R}{E})$, so $t=-\frac{L}{R}\ln(1 - \frac{I R}{E})$.

Step2: Substitute values for part (a)

We are given $E = 17$ volts, $R = 13$ ohms, $L = 1$ henry, and $I = 0.5$ amperes. Substitute these values into the formula $t=-\frac{L}{R}\ln(1 - \frac{I R}{E})$. First, calculate $1-\frac{I R}{E}=1-\frac{0.5\times13}{17}=1-\frac{6.5}{17}=1 - 0.3824 = 0.6176$. Then $t=-\frac{1}{13}\ln(0.6176)$. Since $\ln(0.6176)\approx - 0.4805$, then $t=\frac{0.4805}{13}\approx0.037$ seconds.

Step3: Substitute values for part (b)

We are given $E = 17$ volts, $R = 13$ ohms, $L = 1$ henry, and $I = 1.0$ amperes. Substitute into the formula $t=-\frac{L}{R}\ln(1 - \frac{I R}{E})$. Calculate $1-\frac{I R}{E}=1-\frac{1\times13}{17}=1-\frac{13}{17}=\frac{4}{17}\approx0.2353$. Then $t=-\frac{1}{13}\ln(0.2353)$. Since $\ln(0.2353)\approx - 1.447$, then $t=\frac{1.447}{13}\approx0.111$ seconds.

Answer:

(a) $t = 0.037$ seconds
(b) $t = 0.111$ seconds