Answer
(a) $t=-\frac{5}{13}ln(1-\frac{13}{60}I)$
(b) $0.218s$
Work Step by Step
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 the figure on the following
page. Using calculus, it can be shown that the current
(a) From the original equation, we have $e^{-13t/5}=1-13I/60, -13t/5=ln(1-13I/60)$
so we have $t=-\frac{5}{13}ln(1-\frac{13}{60}I)$
(b) Let $I=2A$, we have $t=-\frac{5}{13}ln(1-2\times\frac{13}{60})=0.218s$
