# 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?

Jul 16, 2018

${V}_{L} = {V}_{B} \cdot {e}^{-} \left(\frac{t R}{L}\right)$

#### Explanation:

${V}_{L} = {V}_{B} \cdot {e}^{-} \left(\frac{t R}{L}\right)$ ....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 ${e}^{-} \left(\frac{t R}{L}\right) = 1$ 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

${V}_{R} = I R$ where $I = {V}_{R} / R \cdot \left(1 - {e}^{-} \left(t \frac{R}{L}\right)\right)$

The graph of these voltages is below.

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 ${e}^{-} \left(\frac{t R}{L}\right) = 1$, so ${V}_{L} = {V}_{B}$ (assuming an ideal voltage source, current will asymptote to infinity if $R = 0$ so this cannot actually happen.)