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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?

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  • \$\begingroup\$ You can just calculate R2 + R1||R3. \$\endgroup\$
    – The Photon
    Commented Jan 21, 2019 at 21:14
  • \$\begingroup\$ But don't forget to include the inductor value in your time constant calculation. \$\endgroup\$
    – The Photon
    Commented Jan 21, 2019 at 21:15

1 Answer 1

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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 \$

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