# RL Transient Analysis confusion

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.

• Check the second approach Aug 4 at 20:21
• @HariKrishna have i made some mistake in it ? Aug 4 at 20:24
• Check my answer , and see if its coming correctly Aug 4 at 20:25
• 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 Aug 4 at 20:55

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