A battery is connected in series with a R = 0.26 Ω resistor and an inductor, as shown in the figure below. The switch is closed at t = 0. The time constant of the circuit is τ = 0.10 s, and the maximum current in the circuit is I = 8.4 A. Find the following.

Question

(a) the emf of the battery V?
(b) the inductance of the circuit
mH ?
(c) the current in the circuit after one time constant has elapsed
A ?

(d) the voltage across the resistor and the voltage across the inductor after one time constant has elapsed
resistor V
inductor V

Answer 1

a) as t--> oo inductor is a short circuit, no reactance
so
V = i at big t * R
V = (8.4)(.26) = 2.14 Volts

b) at t = 0, i = 0, so all voltage across L
2.14 = L di/dt

let i = 8.4 [ 1 - e^-t/tau ]
given tau = .1 s
i = 8.4 [ 1 - e^-10 t ]
di/dt = 8.4 (10 e^-10 t)
di/dt = 84 at t = 0
so
2.14 = L (84)
L = .0255 H

c) when t = tau = .1
i = 8.4 ( 1 - e^-1) = 8.4(.632) = 5.31 amps

d) Vresistor = i R = 5.31*.26 = 1.38 volts
Vinductor = 2.14 - 1.38 = .759 volts