Question

What is the potential difference across a 10 mH inductor if the current through the inductor drops from 150 mA to 50 mA in 10 ms? What is the direction of this potential difference? That is, does the potential increase or decrease along the direction of the current?

Short answer

The potential difference is 100 V.

Step-by-step solution

Given information 

We have given that the self inductance is 10 mH and the current through the inductor drops from 150 mA to 50 mA in 10 ms.

Step 1.

The formula to calculate the potential difference:

$∆V=-L\frac{dI}{dt}$

Here $∆V$= Potential difference

$L$= Self inductance

$dI$= Difference in current

$dt$= Time

substituting the values in formula :

$∆V=-(10\times {10}^{-3}H)\frac{\left(50\times {10}^{-3}A-150\times {10}^{-3}A\right)}{10\times {10}^{-6}s}$

$∆V=-(10\times {10}^{-3})\frac{(50\times {10}^{-3}-150\times {10}^{-3})}{10\times {10}^{-6}}$

$∆V=100V$

Step 2. 

As the current is decreasing, the flux through the center of inductor will also decrease.

To maintain the original flux, induced current will be in the direction of original current. Therefore potential will increase in the direction of current.