Question

The current in an RL circuit drops from 0.89 A to 12 mA in the first second following removal of the battery from the circuit. If L is 13 H, find the resistance R in the circuit.

Answer 1

When the battery is applied to RL ckt, then by applying Kirchoff's law, eqn for Current (I) is given by

  I(t) = Io (1- e-t/k ) .............eqn1

where Io is the peak value of the current, I(t) is the value of current at any instant t, k is the relaxation time

k = L/R ..................eqn2

where L is the inductance of the coil while R is the resistance of coil.

Now, if the battery is removed, then again by applying Kirchoff's law, we have

I(t) = Ioe-t/k ..........eqn 3

Now, t =1 s

Io = 0.89 A

I(t) = 12mA = 12*10-3 A ( 1 mA - 1*10-3A

12 mA = 12 *10-3 A)

L = 13 H

k = L/R = 13/R

Using values of Io, I(t), t, in eqn 3

12*10-3 = 0.89 e-1/k = 0.89e-R/13 (Using value of k )

(12*10-3) / 0.89 = e-R/13

13.483 *10-3 = e-R/13

0.01348 = e-R/13

Taking log of both sides,

log 0.01348 = log e-R/13

log 0.01348 = -(R/13) log e ( log xm = mlogx)

-1.8703 = -(R/13) (log e = 1 )

1.8703 = R/13 (by cancelling out negative sign on both sides)

1.8703*13 = R

R =24.3139 ohm

Hence, in RL ckt, Resistance R is found to be 24.3139 ohm