1
\$\begingroup\$

I've the following circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

And I've to find the time-constant of the voltage across the inductor.


I did the following:

The voltage across the inductor is given by:

$$U_L(t)=100\cdot10^{-3}\cdot\frac{\text{d}}{\text{d}t}\left[\mathcal{L}_\text{s}^{-1}\left\{\frac{\frac{8.6}{\text{s}}}{600+\frac{1}{\frac{1}{1800}+\frac{1}{600+100\cdot10^{-3}\cdot\text{s}}}}\cdot\frac{1800}{1800+600+100\cdot10^{-3}\cdot\text{s}}\right\}_{\left(t\right)}\right]$$

To find the rime constant I solved:

$$\frac{1}{e}\cdot U_L(0)=U_L(t)\space\Longleftrightarrow\space t=\frac{1}{10500}$$

Question: is the answer right? And is there an easier way to the solution?

\$\endgroup\$
  • \$\begingroup\$ You can just calculate R2 + R1||R3. \$\endgroup\$ – The Photon Jan 21 '19 at 21:14
  • \$\begingroup\$ But don't forget to include the inductor value in your time constant calculation. \$\endgroup\$ – The Photon Jan 21 '19 at 21:15
0
\$\begingroup\$

It is correct.

The easier way is the Thevenin equivalent R which is (R1||R3) +R2 = 1050\$\Omega\$ then \$T=L/R= 1 \mathrm{mH}/1050 Ω \$

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.