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=1.0 kΩ and C=4.0 μF, if the current is 6.0 mA at t=4.0 ms", at what time (in ms) will the current be 3.0 mA?

Question

Ace Huff

Physics

19 February, 09:34

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=1.0 kΩ and C=4.0 μF, if the current is 6.0 mA at t=4.0 ms", at what time (in ms) will the current be 3.0 mA?

+3

Answers (1)

Know the Answer?

Angeline Hudson · Answer 1

6.77 ms

Explanation:

τ=RC

= 1 x 10³ x 4 x 10⁻⁶

4 X 10⁻³

t1/2=τ ln 2

= 4 X 10⁻³ X ln 2

= 2.77 x 10⁻³

2.77 ms

Half life = 2.77 ms

Current is 6 mA AT t = 4 ms

and half life is 2.77 ms ie current becomes half after every 2.77 ms

so current will reduce from 6 mA to 3 mA in further 2.77 ms

or at t = 4 + 2.77 = 6.77 ms current will become 3 mA.