An inductor of resistance 5 kO and inductance 2mH is connected to a DC circuit. The time constant is
- 400 ns.
- 10 seconds.
- 40 ms.
Explanation from PART66Online
The inductive time constant is tau = L/R = 0.002H / 5000ohm = 4*10^-7 s = 400 nanoseconds. The small inductance and relatively large resistance give a very short time constant; the 10-second and 40ms answers result from mishandling the kilo-ohm and milli-henry prefixes.
S
Shawez asked · 1 Mar 2015
can anyone please explain? thanks
Community Comments (1)
G
ggmu
8 Aug 2015
The for TIme constant in Inductance circuit is L/R and rest is basic maths 400* 10^-9. Hope it helps
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