I've been may asking the really easy question.

The circuit which was made of the AC source and the coil exists.

\$L:=\$self inductance of the coil.

\$v(t):=v_0*\sin(\omega t)\$ (The voltage between the terminals of the coil.)

\$i(t):=?\$ (The current which flows through the coil.)

We want to know the current which flows through the coil.

\$v_L(t):=-L\frac{di}{dt}\$ (The induced EMF which arises at the coil)

\$-v_0*\sin(\omega t)+L\frac{di}{dt}=0\$ (kirchhoff's voltage law)

\$L\frac{di}{dt}=v_0*\sin(\omega t)\$

\$\frac{di}{dt}=\frac{v_0*\sin(\omega t)}{L}\$

\$\int1\frac{di}{dt} \frac{dt}{1}=\int_{}^{}\frac{v_0*\sin(\omega t)}{L}\frac{dt}{1}\$

\$\int1*di=\frac{v_0}{L}\int_{}^{}\sin(\omega t)dt\$

\$i(t)=\frac{v_0}{L}\frac{(-\cos(\omega t))}{\omega}+C\$ (\$C\$ represents the constant of integration)

\$=\frac{-v_0}{\omega L}\sin(\frac{\pi}{2}-\omega t)+C\$

\$=\frac{v_0}{\omega L}\sin(\omega t -\frac{\pi}{2})+C\$

The below statement is the problem for me.

The textbook substituted \$0\$ to \$C\$ and states that the dc component is assumed as \$0\$ .

Currently I'm unable to get why the value of the constant of integration is zero and the meaning of dc component in this circuit.

Can anyone tell me some idea(s) or the website(s) which describe(s) of it?


1 Answer 1


$$i(t)=v_0/(\omega L) \times (-cos(\omega t) + C)$$ Current cannot change instantly for an AC sinusoidal voltage supply ,so we can use $$i(t)=0 $$at t=0

And we get $$C=1$$ and hence total response would be

$$i(t)=\frac{v_0}{L}\frac{(-\cos(\omega t)+1)}{\omega}$$

$$\frac{v_0}{L\omega}$$ is DC component of the current .

Why we ingnore (DC component)at steady state analysis ?

for a pure inductive circuit we cannot ignore dc component even for steady state analysis but practically pure inductive circuit is impossible and small resistor is always present so the constant of integration for R-L circuit is $$A\exp^{(-R/L)t}$$ And at steady state it decays to zero ,and that why for steady state analysis of pure inductive circuit constant of integration is assumed to be zero

  • \$\begingroup\$ Your math is wrong. To get \$i(0)=0\$, you need \$C=v_0/(\omega L)\$. If \$i(t)\$ was rewritten to \$i(t)=v_0/(\omega L) \times (-cos(\omega t) + C)\$, then your answer \$C=1\$ makes sense. \$\endgroup\$ Mar 10, 2021 at 7:20
  • \$\begingroup\$ @Vicente Cunha thanks I'll edit it \$\endgroup\$
    – user215805
    Mar 10, 2021 at 7:30
  • 1
    \$\begingroup\$ Math still incorrect because you did not adapt the rewritten \$i(t)\$ to the rest of the answer. Your answer still states that \$i(t)=v_0/(\omega L) \times (-cos(\omega t)) + 1\$, which evaluates to \$i(0)=1-v_0/(\omega L)\$. I will contribute with an edit. \$\endgroup\$ Mar 10, 2021 at 8:48
  • \$\begingroup\$ @Vicente Cunha thanks for edit \$\endgroup\$
    – user215805
    Mar 10, 2021 at 9:03

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