Self-Induction — Question 1

Question 1

Self-induced emf and passive terminal voltage have opposite signs

A .0800-H ideal inductor carries current increasing uniformly from .500 A to 2.50 A in .0200 s. Define positive current from terminal a to b and passive voltage $v=V_a-V_b$.

1. Find passive terminal voltage and signed self-induced emf along the current direction.

2. If the current instead decreases at the same rate while remaining positive, find both signs and identify which terminal is at higher potential.

Question 1 - Solution

Self-induced emf and passive terminal voltage have opposite signs

(a) Method and calculation

$$\dot i=(2.50-.500)/.0200=100\,A/s,\quad\boxed{v=L\dot i=+8.00\,V}.$$$$\boxed{\mathcal E_{\mathrm{self}}=-L\dot i=-8.00\,V}.$$

(b) Interpretation and further calculation

$$\dot i=-100\,A/s\Rightarrow\boxed{v=-8.00\,V,\quad\mathcal E_{\mathrm{self}}=+8.00\,V}.$$$$v<0\Rightarrow\boxed{V_b>V_a}.$$

Physical check

The opposing emf is not an additional voltage drop to add on top of L di/dt; these are two descriptions of the same induction effect.