Self-Induction — Question 2

Question 2

A constant current offset changes flux linkage but not the self-induced emf

An ideal constant-L inductor carries $i(t)=I_c+I_0e^{-t/T}$ with positive constants. Define passive voltage along the current reference direction.

1. Find voltage and self-induced emf.

2. Find the voltage-time integral from zero to infinity and explain the role of Ic.

Question 2 - Solution

A constant current offset changes flux linkage but not the self-induced emf

(a) Method and calculation

$$\dot i=-I_0e^{-t/T}/T\Rightarrow\boxed{v=-LI_0e^{-t/T}/T}.$$$$\boxed{\mathcal E_{\mathrm{self}}=+LI_0e^{-t/T}/T}.$$

(b) Interpretation and further calculation

$$\int_0^\infty vdt=L[i(\infty)-i(0)]=\boxed{-LI_0}.$$$$\Lambda=LI_c+LI_0e^{-t/T};\quad\frac d{dt}(LI_c)=0.$$

Physical check

The coil maintains nonzero final linkage LIc while the induced emf tends to zero.