Question

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?

Answer 1

`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.

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.)