A battery is connected in series with a R = 0.34 O resistor and an inductor, as

Question

A battery is connected in series with a R = 0.34 O resistor and an inductor, as shown in the figure below. The switch is closed at t = 0. The time constant of the circuit is t = 0.30 s, and the maximum current in the circuit is I = 7.7 A. Find the following.

a) the emf of the battery (this is 2.618)
V

(b) the inductance of the circuit (this is 102)
mH

(c) the current in the circuit after one time constant has elapsed (this is 4.87)
A

(d) the voltage across the resistor and the voltage across the inductor after one time constant has elapsed (these are the two I can't find)
resistor V
inductor V

For Vr I tried doing Vr=(4.87)(.34) but that isn't right and I have no idea how to find Vi.
Thank you!

Explanation / Answer

Since you belive that the parts a, b and c are correct (and they seem to be)  then le's do  part d)
(d) the voltage across the resistor and the voltage across the inductor after one time constant has elapsed (these are the two I can't find)
resistor V
inductor V

The equation for this circuit is
the emf is equal to the resistor voltag e drop plus voltage drop across the inductor; that is:
Vemf= Vr + Vl for time = time constant a we have
Vr = i(a)R
Vl= Vemf - Vr
finally
Vr= 4.87 x 0.34 = 1.652V (Yes you are correct!  I can't see the circuit bu I assume it is a series ciircuit.)
Vl= 2.618 - 1.652 = 0.964V

Please let me know if it helped.

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