Question

A solenoid having an inductance of 9.13 μH is connected in series with a 1.62 kΩ resistor. (a) If a 16.0 V battery is connected across the pair, how long will it take in seconds for the current through the resistor to reach 82.8% of its final value? (b) What is the current through the resistor at a time t = 1.00τL?

Answer 1

The time constant is given by -

= L / R

= [(9.13 x 10^-6 H) / (1.62 x 10³)]

= 5.635 x 10^-9 sec

(a) An equation for the instantaneous current in RL circuit which will be given by -

I = I₀ (1 - e^-t/)

(0.828) I₀ = I₀ (1 - e^-t/)

(0.828) = 1 - e^-t/

e^-t/ = [1 - (0.828)]

e^-t/ = (0.172)

- (t / ) = ln (0.172)

t = (1.76026)

t = [(1.76026) (5.635 x 10^-9 sec)]

t = 9.92 x 10^-9 sec

t = 9.92 ns

(b) The current through the resistor at a time which will be given by -

I_R = I₀ (1 - e^-t/)

At t = 1 _L, then we have

I_R = (V / R) (1 - e^-1)

I_R = [(16 V) / (1.62 x 10³)] (0.6321)

I_R = [(0.0098765 A) (0.6321)]

I_R = 0.00624 A

I_R = 6.24 mA