Question

An LR circuit contains an ideal 60-V battery, a 42-H inductor having no resistance, a 24-ΩΩ resistor, and a switch S, all in series. Initially, the switch is open and has been open for a very long time. At time t = 0 s, the switch is suddenly closed. How long after closing the switch will the potential difference across the inductor be 24 V?

Answer 1

Answer:

1.6 s

Explanation:

To find the time in which the potential difference of the inductor reaches 24V you use the following formula:

$V_L=V_oe^{-\frac{Rt}{L}}$

V_o: initial voltage = 60V

R: resistance = 24-Ω

L: inductance = 42H

V_L: final voltage = 24 V

You first use properties of the logarithms to get time t, next, replace the values of the parameter:

$\frac{V_L}{V_o}=e^{-\frac{Rt}{L}}\\\\ln(\frac{V_L}{V_o})=-\frac{Rt}{L}\\\\t=-\frac{L}{R}ln(\frac{V_L}{V_o})\\\\t=-\frac{42H}{24\Omega}ln(\frac{24V}{60V})=1.6s$

hence, after 1.6s the inductor will have a potential difference of 24V