Question

Consider an LR circuit like that displayed in Figure 6.6 in Activity 2-1 in your lab manual. You measure the current through the inductor as a function of time, and find that the the current eventually levels off to be 55.05 mA. The voltage of the battery is 6V as displayed.

a) What is the total resistance of the circuit?

b) If the circuit takes 7.52 ms to reach 34.80 mA, what is the value of L for the inductor?

Answer 1

part a)

V= IR

I= 55.05mA

V=6

R= V/I

R=109 ohm

part b)

For an RL circuit we know that ...

i(t) = i(steady state) - i(steady state) e^(-tR/L) .................eq1

And you have given:
i(t=7.52 ms)=34.80 mA
i(steady state)=55.05mA
R=109 ohms

according to eq1
0.03480 = 0.05505 - 0.05505 e^(-0.00752 x 109 / L)
-.02025= -0.05505 e^(-0.00752 x 109 / L)
0.3678= e^(-0.00752 x 109 / L)
ln (0.3678) = ln ( e^(-0.00752 x 109 / L))
ln (0.3678) = -0.00752 x 109 / L
L = -0.00752 x 109 / ln (.3678)
L = -0.00752 x 109 / (-1.0002159668119)
L = 0.8195 H

L= 0.82 H (rounded)