The quantity t1/2=τ ln 2 is called the half-life of an exponential decay, where τ=RC is the time constant in an RC circuit. The current in a discharging RC circuit drops by half whenever t increases by t1/2. For a circuit with R=1 kΩ and C=4 μF, if the current is 6 mA at t=4 ms, at what time (in ms) will the current be 3 mA?

Question

The quantity t1/2=τ ln 2 is called the half-life of an exponential decay, where τ=RC is the time constant in an RC circuit.  The current in a discharging RC circuit drops by half whenever t increases by t1/2.  For a circuit with R=1 kΩ and C=4 μF, if the current is 6 mA at t=4 ms, at what time (in ms) will the current be 3 mA?

rulnick · Answer

tau = RC = 1000(4/1000000) = 4/1000 = 4 ms.

t1/2 = 4 ms (ln 2) = 4ln2 ms.

So the current will be 3mA exactly 4ln2 ms after t=4ms, or at approx. t=6.7726 ms.