Question

What is the potential difference across a 15 mH inductor if the current through the inductor drops from 160 mA to 50 mA in 15 micro seconds?

Answer 1

Concepts and reason

The concepts required to solve the given problem is potential difference in the inductor.

Initially, convert the inductance from mH to H. Later, calculate the change in the current through the inductor. Finally, calculate the potential difference in the inductor.

Fundamentals

The expression for the potential difference in the inductor is as follows:

$V = L\left( {\frac{{dI}}{{dt}}} \right)$

Here, L is the inductance of the inductor, dI is the change in the current in an inductor, and dt is the change in the time.

Convert the inductance of the inductor form mH to H as follows:

$\begin{array}{c}\\L = 15{\rm{ mH}}\left( {\frac{{{{10}^{ - 3}}{\rm{ H}}}}{{1{\rm{ mH}}}}} \right)\\\\ = 15 \times {10^{ - 3}}{\rm{ H}}\\\end{array}$

Firstly, calculate the change in the current as follows:

$\begin{array}{c}\\dI = 160{\rm{ mA}} - 50{\rm{ mA}}\\\\{\rm{ = 110 mA}}\\\end{array}$

Now, substitute 110 mA for dI, $15{\rm{ }}\mu {\rm{s}}$ for dt, and $15 \times {10^{ - 3}}{\rm{ H}}$ for L in the equation $V = L\left( {\frac{{dI}}{{dt}}} \right)$ .

$\begin{array}{c}\\V = \left( {15 \times {{10}^{ - 3}}{\rm{ H}}} \right)\left( {\frac{{110{\rm{ mA}}}}{{15{\rm{ }}\mu {\rm{s}}}}} \right)\\\\ = \left( {15 \times {{10}^{ - 3}}{\rm{ H}}} \right)\left( {\frac{{110{\rm{ mA}}\left( {\frac{{{{10}^{ - 3}}{\rm{A}}}}{{1.00{\rm{ mA}}}}} \right)}}{{15{\rm{ }}\mu {\rm{s}}\left( {\frac{{{{10}^{ - 6}}{\rm{s}}}}{{1.00{\rm{ }}\mu {\rm{s}}}}} \right)}}} \right)\\\\ = 110{\rm{ V}}\\\end{array}$

Ans:

The magnitude of the potential difference across he inductor is equal to 110 V.