Question

RL Circuits 

The current in the RL circuit shown below reaches half its maximum value in 1.75 ms after the switch S1 is thrown. Determine (a) the time constant of the circuit and (b) the resistance of the circuit if L = 250 mH.

Answer 1

In an LR circuit, current at time t is given by:

i = i0*(1 - exp (-t/tau))

i0 = max current =

tau = time constant = ?

Now at time t = 1.75 ms, given that i = i0/2

i0/2 = i0*(1 - exp(-1.75*10^-3/tau)

1/2 = 1 - exp(-1.75*10^-3/tau)

exp(-1.75*10^-3/tau) = 1/2

take logarithm on both sides

tau = -(1.75*10^-3)/ln (1/2)

tau = (1.75*10^-3)/ln (2)

tau = time constant = 2.5247*10^-3 sec

tau = 2.52 ms = 2.52*10^-3 sec

Part B.

time constant of LR circuit is given by:

tau = L/R

R = L/tau

R = 250*10^-3/(2.5247*10^-3)

R = 99.0 ohm = resistance of the circuit

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