The quantity t 1/2 = ln 2 is called the half-life of an exponential decay, where

Question

The quantity t1/2= ln 2 is called thehalf-life of an exponential decay, where =RC is thetime constant in an RC circuit. The current in adischarging RC circuit drops by half whenever t increases byt1/2. For a circuit with R=2 k and C=2 F, ifthe current is 6 mA at t=4 ms, at what time (in ms) will thecurrent be 3 mA? Thanks for the help! The quantity t1/2= ln 2 is called thehalf-life of an exponential decay, where =RC is thetime constant in an RC circuit. The current in adischarging RC circuit drops by half whenever t increases byt1/2. For a circuit with R=2 k and C=2 F, ifthe current is 6 mA at t=4 ms, at what time (in ms) will thecurrent be 3 mA? Thanks for the help! Thanks for the help!

Explanation / Answer

=time constant = RC = 2*103*2*10-6 F = 4 ms so t1/2 = half life = 4 ms * ln 2 = 2.77 ms we know , The current in a discharging RC circuit drops by halfwhenever t increases by t1/2. So it will drop to 3 mA in t = t1/2 =2.77 ms So time at that moment = current time +t = 4 +2.77 = 6.77ms

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