Is this question right in the sense that a time constant can be deduced from this diagram without a voltage supply? If yes, then how do I calculate RxC from this network ? I am very confused.

Edit : my concern and question is whether the teacher should have mentioned an open node where the supply voltage can be assumed, otherwise in some cases the time constant will be 0 ? enter image description here

  • \$\begingroup\$ You only need a source voltage to calculate specific voltages at some time after T=0. \$\endgroup\$ – Peter Smith May 25 at 11:02
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    \$\begingroup\$ Time constants are independent of the voltage source. \$\endgroup\$ – ErikR May 25 at 11:10
  • \$\begingroup\$ I think that you can deduce the time constant by assuming an initial voltage on the capacitor (or an initial current on the inductor) and work out the voltage (current) on the same as a function of time. Time constant should appear in that expression also IIRC. \$\endgroup\$ – AJN May 25 at 11:27
  • \$\begingroup\$ simplifying the left circuit will result into a resistor in parallel with a inductor, and the right network leads to a resistor in parallel with a capacitor. shouldn't they be in a series connection to get a time constant ? Assuming if a voltage source is applied to both terminals of the simplified circuit of the right one then the capacitor will be directly connected to a voltage source and hence the time constant will be ideally 0 ? \$\endgroup\$ – Nafis Ahmed Fahim May 25 at 11:49
  • \$\begingroup\$ Time constants are for series circuits -- i.e. an L and an R in series or a R and a C in series. Also, time constants apply to the current through the series as well as the voltage at the juncture. So perhaps there are some components here you can ignore. \$\endgroup\$ – ErikR May 25 at 11:58

The time constant \$\tau_L \$ can be found with \$\tau_L = \frac{L}{R_{eq}} \$ where \$R_{eq} \$ is the resistance found by taking out the inductor and looking at the equivalent resistance from the terminals.

The almost same argument goes for \$\tau_C \$, but it is found with \$\tau_C=R_{eq}C \$.

  • \$\begingroup\$ I don't know if this is the answer. I think this answer is more correct compared to the other existing answer, where you have to choose two nodes to connect a virtual voltage source. The answer could possibly vary based on the choice of the nodes. This answer however takes into account how the inductor sees the rest of the circuit from its two terminals. \$\endgroup\$ – AJN May 25 at 14:12
  • \$\begingroup\$ The equivalent resistance seen by the capacitor or the inductor would be how the capacitor/inductor relax if it had an initial charge/current. And that relaxation time would be the time constant of the circuit IMO. \$\endgroup\$ – AJN May 25 at 14:18
  • \$\begingroup\$ @AJN Well, the other answer on this question is completely wrong and can be proved to be so with a quick transient simulation. I agree with your understanding of the time constant. \$\endgroup\$ – Carl May 25 at 15:57

The time constant of an R-C series circuit is RC.

The time constant of an L-R series circuit is L/R.

If you attach a voltage source to the top and bottom of each circuit and look at the current drawn you can ignore those parallel branches which do not contain an L-R or R-C series because they will only add a constant (non-time varying) amount to the total current. The rise (or fall) time of the current curve will only depend on the branch containing the L-R or R-C series.

That means the answer for the first one is 20 mH / 2 Ohms = 10 milliseconds.

And for the second it's (20 || 5) F * (4 || 16) Ohms = 4 * 3.2 = 12.8 seconds.

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    \$\begingroup\$ With this logic, if the inductor had an initial current and was left to settle the only resistor that would influence the decay time of the current would be the 2Ohm resistor. That is not right. \$\endgroup\$ – Carl May 25 at 13:33
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    \$\begingroup\$ I didn't down vote, but your solution seems to depend on the choice of nodes to which the virtual or test voltage is applied. By your logic, i can connect a test voltage source directly to the inductor terminals and get time constant of infinity (or zero?) \$\endgroup\$ – AJN May 25 at 14:15
  • \$\begingroup\$ the problem begins with my/student's assumption of supply voltage placement node? then the question becomes open ended and the answers can vary \$\endgroup\$ – Nafis Ahmed Fahim May 25 at 18:14

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