Question

A 10 mH inductor is connected in series with a 10-ohm resistor, a switch and a 6-volt battery.(a) What is the time constant of the circuit?(b) How long after the switch is closed will the current reach 99% of its final value?

Answer 1

Explanation:

It is given that,

Inductance of the inductor, $L=10\ mH=10\times 10^{-3}\ H=10^{-2}\ H$

Resistance of the resistor, R = 10 ohms

(a) Let $\tau$ is the time constant of the circuit. It is given by :

$\tau=\dfrac{L}{R}$

$\tau=\dfrac{10^{-2}}{10}$

$\tau=0.001\ s$

$\tau=1\ ms$

(b) The current equation in RL circuit is given by :

$I'=I(1-e^{\dfrac{-t}{\tau}})$

I' = 0.99 I

$0.99=1-e^{\dfrac{-t}{\tau}}$

$e^{\dfrac{-t}{\tau}}=0.01$

$e^{\dfrac{-t}{1\ ms}}=0.01$

t = 4.6 ms

Hence, this is the required solution.