What is the voltage-time graph for an inductor when the switch closes?

The potential difference between the inductor is the emf of the source when t=o, so how does the p.d. vary until the steady state current?

1 Answer
Jul 16, 2018

# V_L = V_B * e^-((tR)/L)#

Explanation:

# V_L = V_B * e^-((tR)/L) # ....voltage across an inductor

This equation gives voltage (#V_L#)across an inductor, with series resistance #R#, connected across a voltage source #V_B#, at time #t#.

#e^0 = 1#, so:
At time #t=0# the term # e^-((tR)/L) = 1 # so # V_L = V_B #

As time increases the voltage over the inductor decreases exponentially, and the voltage across the series resistance increases exponentially.

Voltage across the resistor:

#V_R =V_B - V_L # and

#V_R = IR # where #I = V_R/R*(1-e^-(tR/L))#

The graph of these voltages is below.

enter image source here

Source https://www.electronics-tutorials.ws/inductor/lr-circuits.html

The blue line represents #V_R# and shows #I# when #R=1# if the vertical voltage and current scales are 1:1

Note that if #R=0# then the term # e^-((tR)/L) = 1 #, so # V_L = V_B # (assuming an ideal voltage source, current will asymptote to infinity if #R=0# so this cannot actually happen.)