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schematic

simulate this circuit – Schematic created using CircuitLab

When I am using Thevenin's circuit and deriving inductor currents equations i am getting different Time constant \$ \tau\$ from what I am getting when I use Node Current analysis.

Here is my work for Thevenin's Circuit:-

\$from\ KVL :\$

\$V_{th} = i*R_{th} + L*\frac{di(t)}{dt}\$

\$\frac{V_{th}}{L}\ = i*\frac{R_{th}}{L}\ + \frac{di(t)}{dt}\$

Now after solving the homogeneous equation I get

\$i(t)= A*e^{-\frac{t}{\tau}\ \hspace{35pt} Where\: A\: is\: Constant}\$

\$ \tau = \frac{L}{R_{Th}}\ = \frac{0.8}{8}\ = 0.1s \$

This is what I am getting from Thevenin's equivalent circuit now Look at Original Circuit Equation

\$from\ KCL\: at\ node\: A :\$

\$ \frac{V_s-v}{R1}\ + \frac{0-v}{R2}\ - i = 0 \$

\$ \frac{V_s}{10}\ - v*(\ \frac{1}{10}\ +\frac{1}{40}\ ) - i =0 \$

\$ 8*v+i = \frac{V_s}{10}\ \$

\$ now\qquad v = L*\frac{di(t)}{dt}\ \$

\$ \frac{di(t)}{dt}\ + \frac{i}{8L}\ = \frac{V_s}{80L}\ \$

Now solving homogeneous equation

\$ \frac{di(t)}{dt}\ = - \frac{i}{8L}\ \$

\$i(t)= A*e^{-\frac{t}{\tau}\ \hspace{35pt} Where\: A\: is\: Constant}\$

\$ \tau = L*8 = 0.8*8 = 6.4s \$

Now here i am getting different \$ \tau \$ from what i get in Thevenin's circuit...also source is just constant. Please tell me where I'm making mistake.

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  • \$\begingroup\$ Check the second approach \$\endgroup\$ Aug 4 at 20:21
  • \$\begingroup\$ @HariKrishna have i made some mistake in it ? \$\endgroup\$
    – fpsshubham
    Aug 4 at 20:24
  • \$\begingroup\$ Check my answer , and see if its coming correctly \$\endgroup\$ Aug 4 at 20:25
  • \$\begingroup\$ The current flowing through the inductor at t->infinity is 1A .8Ω is the resistance seen by the inductor.So the final equation for inductor current is:IL(t)=1(1-e^(-10t)) so the time costant is 0.1s \$\endgroup\$
    – Miss Mulan
    Aug 4 at 20:55
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Your first approach is correct. Second one, you took the KCL at node A (correct) but then two lines from there its V/8 but you wrote it V*8

PS: \$-V(\frac{1}{10} + \frac{1}{40}) = -\frac{V}{8} \$ ; not -V8!!!

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  • \$\begingroup\$ Ohhh ok i understand all this time i was inversing 1/8 because i was thinking about parallel resistance!!! \$\endgroup\$
    – fpsshubham
    Aug 4 at 20:29
  • \$\begingroup\$ Yea it should be matching now \$\endgroup\$ Aug 4 at 20:31

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