How is the time constant for an LR circuit is determined in the case of charging, where one resistor is connected in parallel and another resistor is connected series to an inductor. 
\$\begingroup\$
\$\endgroup\$
4
-
\$\begingroup\$ Show the circuit you have in mind, because it depends on where the parallel resistor is located (in parallel with the source or in parallel with the inductor). If it is the latter, then Andy’s answer below is what you want. \$\endgroup\$– relayman357Commented Nov 15, 2020 at 17:43
-
\$\begingroup\$ Sir i have added it please check now \$\endgroup\$– SudharsanCommented Nov 15, 2020 at 19:04
-
\$\begingroup\$ Please check now \$\endgroup\$– SudharsanCommented Nov 15, 2020 at 19:05
-
\$\begingroup\$ Thanks for adding the picture. It is clear now, see my answer. \$\endgroup\$– relayman357Commented Nov 15, 2020 at 19:28
Add a comment
|
1 Answer
\$\begingroup\$
\$\endgroup\$
4
Since your upper resistor is in parallel with the ideal voltage source, it will not be part of the time constant for the inductor transient. You can ignore it as if it’s not even there.
answered Nov 15, 2020 at 19:27
-
\$\begingroup\$ Sir why we are neglecting the above resistance? \$\endgroup\$ Commented Nov 15, 2020 at 19:48
-
\$\begingroup\$ But what u have said is the same on my book solution \$\endgroup\$ Commented Nov 15, 2020 at 19:49
-
\$\begingroup\$ Sir.... please post an image of what circuit mr. Andy has explained.... iam confused with that \$\endgroup\$ Commented Nov 15, 2020 at 19:50
-
\$\begingroup\$ You are welcome. Andy (I think) was thinking your upper resistor was in parallel with just the inductor. \$\endgroup\$ Commented Nov 15, 2020 at 20:15